Circular Motion and Universal Gravitation

Question 1

Question 1: 1. The following historical facts about the application of Kepler's laws and the law of gravity to t...

1. The following historical facts about the application of Kepler's laws and the law of gravity to the study of astronomy are correct ( )

A. A. The area swept by Mars in line with the Sun is equal to the area swept by Jupiter in line with the Sun for the same period of time
B. B. The cube of the ratio of the periods of rotation of Mars and Jupiter is equal to the square of the ratio of the half-length axes of their orbits.
C. C. Neptune was discovered by the British physicist Cavendish after a lot of calculations, and has been called "the planet on the tip of a pen.
D. D. The discovery of Neptune demonstrates how theory can be a great guide to practice, and the method of "calculation, prediction and observation" used guided the search for new celestial bodies.

Answer

Answer: D

Solution: A. According to Kepler's second law, the area swept by the line between Mars and the Sun is not equal to the area swept by the line between Jupiter and the Sun at the same time, so A is wrong; B. According to Kepler's third law, the square of the ratio of the periods of rotation of Mars and Jupiter is equal to the cube of the ratio of the half-length axes of their orbits, so B is wrong; C. Neptune was discovered after a lot of calculations using the law of gravity, but not by Cavendish after a lot of calculations using the law of gravity, so C is wrong; D. Neptune's discovery process fully demonstrates the theory of the practice of the great guiding role of the method of "calculation, prediction and observation" to guide people to find new celestial bodies, so D is correct;

Question 2

Question 2: 2. The following physical quantities are scalars (

2. The following physical quantities are scalars (

A. A. displacement (vector)
B. B. accomplishment
C. C. strength
D. D. tempo

Answer

Answer: B

Solution: Work only has magnitude and no direction and is a scalar; displacement, force and velocity have both magnitude and direction and are vectors, so it can be seen that $B$ is a scalar and $A , C , D$ is a vector;

Question 3

Question 3: 3. The scientist who first measured the gravitational constant more accurately by experiment was

3. The scientist who first measured the gravitational constant more accurately by experiment was

A. A. Hooker, England.
B. B. Newton of England.
C. C. Galileo of Italy
D. D. Britain's Cavendish.

Answer

Answer: D

Solution: The value of the gravitational constant was derived in 1789 by Cavendish using a torsion scale he invented.

Question 4

Question 4: 4. The units of the following physical quantities are described in terms of the base units of the In...

4. The units of the following physical quantities are described in terms of the base units of the International System of Units (SI), and the descriptions are correct

A. A. The dynamic friction factor $\mu$ is in units of $\mathrm { kg } / \mathrm { s } ^ { 2 }$.
B. B. The gravitational constant $G$ has units of $\mathrm { m } ^ { 2 } / \left( \mathrm { kg } \cdot \mathrm { s } ^ { 2 } \right)$
C. C. The spring strength factor $k$ is in units of $\mathrm { N } / \mathrm { m }$.
D. D. The static power constant $k$ has units of $k g \cdot m ^ { 3 } / \left( s ^ { 4 } \cdot A ^ { 2 } \right)$.

Answer

Answer: D

Solution: A. According to $\mu = \frac { f } { F _ { \mathrm { N } } }$, the kinetic friction factor $\mu$ has the unit $1 , \mathrm {~A}$; B. According to the law of gravity $F = G \frac { m _ { 1 } m _ { 2 } } { r ^ { 2 } }$, we can get $$ G = \frac { F r ^ { 2 } } { m _ { 1 } m _ { 2 } } $$ B. According to the law of gravity $F = G \frac { m _ { 1 } m _ { 2 } } { r ^ { 2 } }$, $$ G = \frac { F r ^ { 2 } } { m _ { 1 } m _ { 2 } } $$ is $$ G = \frac { F r ^ { 2 } } { m _ { 1 } m _ { 2 } } $$, and the unit of force is $1 \mathrm {~N} = 1 \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s } ^ { 2 }$, the fundamental unit of the gravitational constant G is $\mathrm { m } ^ { 3 } / \mathrm { kg } \cdot \mathrm { s } ^ { 2 }$; C. The coefficient of strength $k$ of a spring has the unit $\mathrm { N } / \mathrm { m }$, but N is the derived unit, C is wrong; D. According to $F = \frac { k Q q } { r ^ { 2 } }$, the coefficient of strength of a spring is $F = \frac { k Q q } { r ^ { 2 } }$. $$ k = \frac { F r ^ { 2 } } { Q q } $$ From $q = I t$ and $F = m a$, we have the following joint simplification The electrostatic constant $k$ has units of $\mathrm { kg } \cdot \mathrm { m } ^ { 3 } / \left( \mathrm { s } ^ { 4 } \cdot \mathrm {~A} ^ { 2 } \right)$, and D is correct.

Question 5

Question 5: 5. If two satellites are in uniform circular motion around the Earth, the one that is closer to the ...

5. If two satellites are in uniform circular motion around the Earth, the one that is closer to the surface is

A. A. Smaller angular velocity
B. B. Shorter operating cycles
C. C. Smaller linear velocity
D. D. Less centripetal acceleration

Answer

Answer: B

Solution: A. Based on $$ G \frac { M m } { r ^ { 2 } } = m \omega ^ { 2 } r $$ it can be obtained that $$ \omega = \sqrt { G \frac { M } { r ^ { 3 } } } $$ It can be seen that the angular velocity of satellites closer to the ground is larger, A is wrong; B. According to $$ \mathrm { G } \frac { M m } { r ^ { 2 } } = m \frac { 4 \pi ^ { 2 } } { T ^ { 2 } } r ^ { 2 } $$ B. According to $$ \mathrm { G } \frac { M m } { r ^ { 2 } } = m \frac { 4 \pi ^ { 2 } } { T ^ { 2 } } r ^ { 2 } $$ $$ T = \sqrt { \frac { 4 \pi ^ { 2 } } { G M } r ^ { 3 } } $$ It can be seen that satellites closer to the ground have shorter operation periods, B is correct; C. According to C. According to $$ \mathrm { G } \frac { M m } { r ^ { 2 } } = m \frac { v ^ { 2 } } { r } $$ can be obtained $$ v = \sqrt { \frac { G M } { r } } $$ It can be known that satellites closer to the ground have larger linear velocities, C is wrong; D. According to $$ G \frac { M } { r ^ { 2 } } = a $$ it can be known that satellites closer to the ground have a greater centripetal acceleration, D is wrong;

Question 6

Question 6: 6. In the process of mankind's understanding of the laws of motion of matter, many physicists boldly...

6. In the process of mankind's understanding of the laws of motion of matter, many physicists boldly conjecture, the courage to question, and achieved brilliant achievements. The following description of the scientists and their contributions, the correct one is

A. A. Kepler studied the astronomical observations of the Diya and proposed that the planets move in a uniform circular motion around the sun
B. B. Cavendish arrived at the value of $G$ more accurately by measuring the gravitational force between lead balls
C. C. Newton discovered the law of gravity, from which it can be deduced that the Earth and Mars are equal in area swept by the line to the Sun at the same time
D. D. The "planet found under the tip of a pen" is Uranus, calculated by the law of gravity.

Answer

Answer: B

Solution: A. Kepler studied the astronomical observations of the Diya and proposed that the planets move in elliptical motion around the Sun, so A is wrong; B. Cavendish used the torsion scale experiment, by measuring the gravitational force between lead balls, more accurately arrived at the value of $G$, so B is correct; C. Newton discovered the law of gravity, according to Kepler's second law can be known, the same planet moving around the sun, the planet and the sun line at the same time swept across the area is equal, when the earth and mars at the same time and the sun line swept across the area is not equal, so the C error; D. calculated by the law of gravity "found under the tip of the planet" for Neptune, so D error.

Question 7

Question 7: 7. With respect to the description of curvilinear motion, the following statements are correct

7. With respect to the description of curvilinear motion, the following statements are correct

A. A. If the combined force on an object is constant, the object must move in a straight line
B. B. The combined external force on an object in curved motion can be zero
C. C. The combined external force on an object in curved motion points in the direction of the inside of the curve
D. D. The speed and acceleration of an object in uniform circular motion are constant, so it is also called uniformly variable motion.

Answer

Answer: C

Solution: A. When the combined force on an object is constant, the object can move in a straight line or in a curved line, only when the direction of the combined force and the direction of the velocity are in the same line can the object move in a straight line, so A is wrong; B. Curve motion acceleration is not zero, the force must not be zero, so B error; C. The direction of the force on the object in curved motion points to the inside of the curve, so C is correct; D. Uniform circular motion speed and acceleration size is unchanged, but the direction of change, is non-uniformly variable motion, so D error.

Question 8

Question 8: 8. For the expression $F = G \frac { m _ { 1 } m _ { 2 } } { r ^ { 2 } }$ for the law of gravity, on...

8. For the expression $F = G \frac { m _ { 1 } m _ { 2 } } { r ^ { 2 } }$ for the law of gravity, one of the following statements is correct

A. A. As $r$ tends to zero, the force of gravity tends to infinity
B. B. The $G$ in the formula is the gravitational constant, which is measured experimentally and not artificially specified
C. C. If $m _ { 1 } > m _ { 2 }$, then the gravitational force on ${ } _ { 1 }$ is greater than the gravitational force on ${ } ^ { m _ { 2 } }$.
D. D. The gravitational forces on ${ } ^ { m _ { 1 } }$ and $^ { m _ { 2 } }$ are equal in magnitude and are a balanced pair.

Answer

Answer: B

Solution: A. When $r$ tends to zero, the size and shape of the two objects can not be ignored, that is, can not be regarded as a mass, the law of gravity is established on the premise that the two objects can be regarded as a mass, so you can not simply use mathematical methods to discuss, mistakenly believe that the force of gravity is infinite, option A error; B. The formula $G$ for the gravitational constant, is an eternal constant, it is measured by Cavendish experiment, option B is correct; CD. regardless of the mass of the two objects who is greater and who is smaller, the gravitational force between the objects is always a pair of interacting forces, equal in size, acting on the two objects, options CD are wrong.

Question 9

Question 9: 9. As shown in the figure, astronomers observed a planet and the Earth in the same orbital plane aro...

9. As shown in the figure, astronomers observed a planet and the Earth in the same orbital plane around the Sun to do uniform circular motion, and the planet's orbital radius than the Earth's orbital radius is smaller than the Earth's orbit radius, the Earth and the center of the Sun and the Earth and the planet's connecting line of the angle is called the Earth's view of the planet's view (referred to as the angle of view). The maximum angle of view of the planet is known to be $\theta$. Then the rotation of the Earth and the planet around the Sun ![](/images/questions/phys-circular-motion/image-001.jpg)

A. A. The angular velocity ratio is $\sqrt { \sin ^ { 3 } \theta }$
B. B. The linear velocity ratio is $\sqrt [ 3 ] { \sin \theta }$
C. C. The centripetal acceleration ratio is $\sin \theta$
D. D. The centripetal force ratio is $\sin ^ { 2 } \theta$

Answer

Answer: A

Solution: ABC.The planets and the Earth move in uniform circular motion around the Sun, with gravity providing the centripetal force, and there are $$ \frac { G M m } { r ^ { 2 } } = m \frac { v ^ { 2 } } { r } m \omega ^ { 2 } r = m a _ { n } $$ Solution. $$ v = \sqrt { \frac { G M } { r } } , \omega = \sqrt { \frac { r ^ { 3 } } { G M } } , a _ { n } = \frac { G M } { r ^ { 2 } } $$ From the geometric relationship, the ratio of the orbital radii of the planet to the Earth is $$ \frac { r _ { \text {行星 } } } { r _ { \text {地球 } } } = \sin \theta $$ Then the ratio of angular velocity, linear velocity, and centripetal acceleration of the Earth to that of the planet rotating around the Sun are respectively $$ \frac { \omega _ { \text {行星 } } } { \omega _ { \text {地球 } } } = \sqrt { \sin ^ { 3 } \theta } , \frac { v _ { \text {行星 } } } { v _ { \text {地球 } } } = \sqrt { \sin \theta } , \frac { a _ { \text {n行星 } } } { a _ { \text {n地球 } } } = \sin ^ { 2 } \theta $$ Therefore, A is correct; B is wrong; C is wrong; D. The centripetal force of the Earth and the planets rotating around the Sun is $$ F _ { n } = \frac { G M m } { r ^ { 2 } } $$ Because the mass relationship between the two is unknown, it cannot be determined. Therefore, D is incorrect.

Question 10

Question 10: 10. Two perpendicular co-point forces ${ } ^ { F _ { 1 } }$ and ${ } ^ { F _ { 2 } }$ act on an obje...

10. Two perpendicular co-point forces ${ } ^ { F _ { 1 } }$ and ${ } ^ { F _ { 2 } }$ act on an object and cause the object to undergo a displacement in which the forces ${ } ^ { F _ { 1 } }$ do work on the object of 4 J. The force ${ } ^ { F _ { 2 } }$ does 3 J. The total work done by the combined forces ${ } ^ { F _ { 1 } }$ and ${ } ^ { F _ { 2 } }$ is 3 J. Then the total work done on the object is The force ${ } ^ { F _ { 2 } }$ does work on the object by 3 J. The total work done on the object by the force ${ } ^ { F _ { 1 } }$ and the force ${ } ^ { F _ { 2 } }$ together is

A. A. 0
B. B. 1J
C. C. 5 J
D. D. 7 J

Answer

Answer: D

Solution: Work is a scalar quantity, the work done by the combined force is the sum of the work done by the individual component forces, so the total work done by the combined forces $F _ { 1 }$ and $F _ { 2 }$ on the object is $$ W _ { \text {合 } } = W _ { 1 } + W _ { 2 } = 7 \mathrm {~J} $$

Question 11

Question 11: 11. The following statements about the history of physics are correct

11. The following statements about the history of physics are correct

A. A. Newton was the first to use a combination of experimentation and reasoning to conclude that force is not responsible for keeping objects in motion
B. B. Cavendish is credited with "weighing the Earth" for his measurement of the gravitational constant $G$.
C. C. Aristotle believed that the speed with which an object falls is independent of its weight.
D. D. Kepler discovered the law of gravity by analyzing astronomical observations in the Tiguya

Answer

Answer: B

Solution: A. Galileo was the first to use a combination of experiments and reasoning to find that force is not the cause of maintaining the motion of an object. Therefore, A is wrong; B. Cavendish measured the gravitational constant $G$ with a torsion scale and was known as the man who could "weigh the earth", so B is correct; C. Aristotle believed that the falling speed of an object is related to its weight, so C is wrong; D. Kepler by analyzing the astronomical observations of the valley, put forward Kepler's three laws, Newton discovered the law of gravity, so D error.

Question 12

Question 12: 12. A globular nebula of uniform mass distribution and radius $R$ has a surface gravitational accele...

12. A globular nebula of uniform mass distribution and radius $R$ has a surface gravitational acceleration of $g$. The radius of the nebula changes to $2 R$ due to thermal expansion, and if the mass remains constant and uniformly distributed during this process, then the nebula changes to $2 R$ with $g$. Neglecting the rotation of the nebula, the acceleration of gravity on the surface of the changed nebula is ( )

A. A. $\frac { g } { 4 }$
B. B. $\frac { g } { 8 }$
C. C. $\frac { g } { 16 }$
D. D. $\frac { g } { 64 }$

Answer

Answer: A

Solution: Gravity of objects on the surface of the nebula is equal to the force of gravity. $$ G \frac { M m } { R ^ { 2 } } = m g $$ Therefore $$ g = G \frac { M } { R ^ { 2 } } $$ When the radius becomes twice the original, the acceleration of gravity is $0.25 g$.

Question 13

Question 13: 13. The radius of the Earth is R and the acceleration of gravity at the surface of the Earth is g. I...

13. The radius of the Earth is R and the acceleration of gravity at the surface of the Earth is g. If the acceleration of gravity at a certain height from the ground is $\frac { g } { 6 }$, then the height from the ground at that point is

A. A. $( \sqrt { 3 } - 1 ) _ { R }$
B. B. $( \sqrt { 6 } - 1 ) _ { R }$
C. C. $\sqrt { 6 } R$
D. D. $5 R$

Answer

Answer: B

Solution: Let the mass of the Earth be M , the mass of the object be m , and the height of the object from the ground be h . According to the gravitational approximation According to the approximation of gravity, there is $\mathrm { mg } = \mathrm { G } \frac { M m } { R ^ { 2 } }$ on the surface of the Earth, and there is $m \frac { g } { 6 } = G \frac { M m } { ( R + h ) ^ { 2 } }$ at a height of h. The solution is: $\mathrm { h } = \left( { } ^ { \sqrt { 6 } } - 1 \right) \mathrm { R }$, so B is correct and ACD is wrong. Therefore, B is correct and ACD is wrong. [Eyes on] the key to this question to know the relationship between gravity and gravity, clear in the case of the Earth's rotation is not considered gravity is approximately equal to gravity, know the relationship between the acceleration of gravity and height, and can be used to analyze practical problems.

Question 14

Question 14: 14. The following statements about the work done by a force on an object are correct ( )

14. The following statements about the work done by a force on an object are correct ( )

A. A. Work is a vector quantity, its positive or negative indicates direction
B. B. If the force does not do work on the object, then the object must be at rest
C. C. A force does negative work on an object, or it can be said that the object does work to overcome that force
D. D. The work done by pushing the box with a force of 300 N must be more than the work done by pushing the box with a constant force of 100 N.

Answer

Answer: C

Solution: A. Work is a scalar quantity, its positive or negative indicates the direction of energy transfer, so A error; B. force does not do work on the object, the object is not necessarily at rest, such as uniform linear motion combined external force does not do work, so B error; C. Force on the object to do negative work, you can also say that the object to overcome the force to do work, so C is correct; D. According to $$ W = F x \cos \theta $$ D. According to $$ W = F x \cos \theta $$, the amount of work done by the force on the object is related to the magnitude of the force, the magnitude of the displacement, and the cosine of the angle between the force and the displacement, so D is wrong.

Question 15

Question 15: 15. three horizontally placed three different materials made of round wheel $A , B , C$, with non-sl...

15. three horizontally placed three different materials made of round wheel $A , B , C$, with non-slip belt connected, as shown in the figure (top view), the three round wheel radius ratio is $R _ { A } : R _ { B } : R _ { C } = 3 : 2 : 1$, when the active wheel $C$ rotate at a uniform speed, in the edge of the three wheels were placed on the When the active wheel $C$ is rotating at a uniform speed, a small object $P$ (which can be regarded as a mass), $P$ is able to be relatively stationary on the edges of the three wheels, and the maximal static friction on the object $P$ is equal to the sliding friction, and the small object $P$ is equal to the sliding friction, and the maximal static friction between the object $A , B , C$ and wheel $\mu _ { A } , \mu _ { B } , \mu _ { C } , A , B , C$ is equal to the sliding friction. ] and the wheel $A , B , C$ have a kinetic friction factor of $\mu _ { A } , \mu _ { B } , \mu _ { C } , A , B , C$ between the contact surfaces, respectively, and the angular velocities of rotation of the three wheels are $\omega _ { A } , \omega _ { B } , \omega _ { C }$, respectively, and () ![](/images/questions/phys-circular-motion/image-002.jpg)

A. A. $\mu _ { A } : \mu _ { B } : \mu _ { C } = 2 : 3 : 6$
B. B. $\mu _ { A } : \mu _ { B } : \mu _ { C } = 6 : 3 : 2$
C. C. $\omega _ { A } : \omega _ { B } : \omega _ { C } = 1 : 2 : 3$
D. D. $\omega _ { A } : \omega _ { B } : \omega _ { C } = 6 : 3 : 2$

Answer

Answer: A

Solution: The small object $P$ is only subject to maximum static friction horizontally, which provides the centripetal force, so the centripetal acceleration $a = \mu g$ , and $a = \frac { v ^ { 2 } } { R } , A B C$ the linear velocity at the edge of the three wheels is of the same magnitude so that $\mu \propto \frac { 1 } { R }$ , and so that $\mu _ { A } : \mu _ { B } : \mu _ { C } = 2 : 3 : 6$ , and so $v = R \omega$ is the same. $\mu _ { A } : \mu _ { B } : \mu _ { C } = 2 : 3 : 6$; from $v = R \omega$, $\omega \propto \frac { 1 } { R }$, so $\omega _ { A } : \omega _ { B } : \omega _ { C } = 2 : 3 : 6$, BCD is wrong A is correct.

Question 16

Question 16: 16. On October 18, 2006, the world's first female "space tourist" Ansari took the Soyuz spacecraft a...

16. On October 18, 2006, the world's first female "space tourist" Ansari took the Soyuz spacecraft and successfully flew into space, and she stayed in the International Space Station (ISS) for nine days, where she participated in a number of important experiments of the European Space Agency (ESA). The International Space Station is a place where all kinds of experiments are carried out and the instruments used have to be selected, the following instruments can still be used in the space station

A. A. mercury barometer
B. B. scales (weigh things)
C. C. pendulum clock
D. D. sensitive ammeter

Answer

Answer: D

Solution: A. The mercury barometer relies on atmospheric pressure to support the height of the mercury column, and in a weightless environment the mercury cannot form a liquid column and cannot be used, so A is wrong; B. The balance measures mass by comparing the gravity of objects and weights, and cannot be used in weightlessness when the gravity is zero, so B is wrong; C. The pendulum clock relies on gravity to provide a return force to make the pendulum pendulum, weightlessness pendulum can not be periodic motion, so C error; D. Sensitive ammeter through the magnetic field on the current force to drive the needle deflection, has nothing to do with gravity, so D is correct.

Question 17

Question 17: 17. A spacecraft arrives at a planet (which has no rotational motion), approaches the equatorial sur...

17. A spacecraft arrives at a planet (which has no rotational motion), approaches the equatorial surface of the planet at a speed of $v$ and travels at a constant velocity, the period of the motion is measured to be $T$, and the constant of gravitational attraction is known to be $G$, which is obtained as

A. A. The radius of the planet is $\frac { v T } { \pi }$
B. B. The average density of the planet is $\frac { 3 \pi } { G T ^ { 2 } }$
C. C. The planet has a mass of $\frac { v ^ { 3 } T } { \pi G }$
D. D. The acceleration of gravity on the surface of the planet is $\frac { 2 \pi v } { G T }$

Answer

Answer: B

Solution: A. Based on $$ v T = 2 \pi R $$ The radius of the planet can be obtained as $$ R = \frac { v T } { 2 \pi } $$ A Error ; BC. According to $$ \begin{aligned} G \frac { M m } { R ^ { 2 } } & = m \frac { 4 \pi ^ { 2 } } { T ^ { 2 } } R \\ \rho & = \frac { M } { \frac { 4 } { 3 } \pi R ^ { 3 } } \end{aligned} $$ the mass of the planet can be obtained from $$ \begin{aligned} G \frac { M m } { R ^ { 2 } } & = m \frac { 4 \pi ^ { 2 } } { T ^ { 2 } } R \\ \rho & = \frac { M } { \frac { 4 } { 3 } \pi R ^ { 3 } } \end{aligned} $$. $$ M = \frac { v ^ { 3 } T } { 2 \pi G } $$ The average density is $$ \rho = \frac { 3 \pi } { G T ^ { 2 } } $$ B is correct and C is incorrect ; D. According to $$ G \frac { M m } { R ^ { 2 } } = m g $$ The acceleration of gravity on the surface of the planet is $$ g = \frac { 2 \pi v } { T } $$ D Error.

Question 18

Question 18: 18. At 1:30 on December 2, 2013, the Chang'e-3 lunar probe was launched on a Long March 3B rocket. T...

18. At 1:30 on December 2, 2013, the Chang'e-3 lunar probe was launched on a Long March 3B rocket. The satellite is in uniform circular motion in orbit at an altitude of $h$ from the surface of the Moon, with a period of $T$, and eventually achieves a soft landing on the surface of the Moon. If the $R$ represents the radius of the Moon, and the gravitational constant is $G$, ignoring the rotation of the Moon and the influence of the Earth on satellites, the following statements are not correct The magnitude of the velocity is $\frac { 4 \pi ^ { 2 } ( R + h ) ^ { 3 } } { \mathrm { G } T ^ { 2 } }$

A. A. The mass of the Moon is $\frac { 4 \pi ^ { 2 } ( R + h ) ^ { 3 } } { G T ^ { 2 } }$
B. B. The Moon's first cosmic velocity is $\frac { 2 \pi } { T } \sqrt { \frac { ( R + h ) ^ { 3 } } { R } }$
C. C. The centripetal acceleration of Chang'e-3 while orbiting the Moon is $\frac { 4 \pi ^ { 2 } R } { T ^ { 2 } }$
D. D. The addition of an object in free fall on the surface of the Moon

Answer

Answer: C

Solution: A. Providing centripetal force based on gravity $$ G \frac { M m } { ( R + h ) ^ { 2 } } = m \frac { 4 \pi ^ { 2 } ( R + h ) } { T ^ { 2 } } $$ Solution: $$ M = \frac { 4 \pi ^ { 2 } ( R + h ) ^ { 3 } } { G T ^ { 2 } } $$ A The statement is correct; B. The first cosmic velocity at the surface of the Moon is $$ G \frac { M m } { R ^ { 2 } } = m \frac { v ^ { 2 } } { R } $$ The joint solution is $$ v = \frac { 2 \pi \sqrt { R ( R + h ) ^ { 3 } } } { T R } $$ B. The statement is correct; C. The centripetal force is provided by gravity. $$ G \frac { M m } { ( R + h ) ^ { 2 } } = m \frac { 4 \pi ^ { 2 } ( R + h ) } { T ^ { 2 } } = m a $$ Solution: $$ a = \frac { 4 \pi ^ { 2 } ( R + h ) } { T ^ { 2 } } $$ C. The statement is wrong; D. According to the golden substitution $G M = g R ^ { 2 }$ and $M = \frac { 4 \pi ^ { 2 } ( R + h ) ^ { 3 } } { G T ^ { 2 } }$, the acceleration of gravity on the surface of the Moon is obtained by the conjunction of $G M = g R ^ { 2 }$ and $M = \frac { 4 \pi ^ { 2 } ( R + h ) ^ { 3 } } { G T ^ { 2 } }$. $$ g = \frac { 4 \pi ^ { 2 } ( R + h ) ^ { 3 } } { G T ^ { 2 } } $$ D is correct. This question is incorrect, so choose C.

Question 19

Question 19: 19. With respect to the first cosmic velocity, the following statements are true

19. With respect to the first cosmic velocity, the following statements are true

A. A. The first cosmic velocity is the maximum speed at which an artificial Earth satellite can be launched
B. B. The first cosmic velocity is the minimum speed at which artificial Earth satellites orbit around
C. C. The Earth's first cosmic velocity is determined by the Earth's mass and radius
D. D. The first cosmic velocity is the speed at which a geostationary satellite orbits.

Answer

Answer: C

Solution: A. The first cosmic velocity is the minimum speed at which an artificial Earth satellite can be launched; B. According to ${ } ^ { v } = \sqrt { \frac { G M } { r } }$, it is known that the first cosmic velocity is the maximum speed at which an artificial Earth satellite can orbit; C. According to ${ } ^ { v } = \sqrt { \frac { G M } { r } }$, it is known that the first cosmic velocity of the Earth is determined by the mass and radius of the Earth, option C is correct; D. The first cosmic velocity is not the speed at which the geostationary satellites orbit around the Earth, which is greater than that of the geostationary satellites, option D is wrong.

Question 20

Question 20: 20. The magnitude of the Earth's gravitational force on an object is equal to the object's gravitati...

20. The magnitude of the Earth's gravitational force on an object is equal to the object's gravitational force on the Earth, but we always see the object falling towards the Earth and the Earth does not move toward the object, this is because

A. A. The law of gravity does not apply to the earth and objects
B. B. Newton's third law does not apply to the earth and objects
C. C. Using an object on Earth as a frame of reference, you can't see the Earth moving toward the object, but if you use the Sun as a frame of reference, you can see the Earth moving toward the object
D. D. The Earth's mass is so great that it produces so little acceleration that even with the Sun as a reference point, the Earth cannot be seen moving toward an object

Answer

Answer: D

Solution: AB. The law of gravity applies to the Earth and objects, and the force of gravity between two objects is also a pair of action and reaction forces, also following Newton's third law, so AB is wrong; CD.Because the mass of the Earth is too large, the resulting acceleration is very small, even with the Sun as a reference, you can't see the Earth moving toward the object, so C is wrong and D is correct.

Question 21

Question 21: 21. As shown in the figure, the $A , B$ wheel is driven by a belt, the $A , C$ wheel is driven by fr...

21. As shown in the figure, the $A , B$ wheel is driven by a belt, the $A , C$ wheel is driven by friction, radius $R _ { A } = 2 R _ { B } = 3 R _ { C }$, and all contact surfaces do not slip, then the ratio of linear and angular velocities at the edge points of the three wheels of $A , B , C$ are at the edge points of the three wheels are ![](/images/questions/phys-circular-motion/image-003.jpg) $A \cdot \mathrm { v } _ { \mathrm { A } } : \mathrm { v } _ { \mathrm { B } } : \mathrm { v } _ { \mathrm { C } } = 1 : 2 : 3 , \omega _ { \mathrm { A } } : \omega _ { \mathrm { B } } : \omega _ { \mathrm { C } } = 3 : 2 : 1$

A. A. As shown in the figure, the $A , B$ wheel is driven by a belt, and the $A , C$ wheel is driven by friction, with the radius $R _ { A } = 2 R _ { B } = 3 R _ { C }$ , and none of the contact surfaces are slipping, then $A , B , C$
B. B. $\mathrm { v } _ { \mathrm { A } } : \mathrm { v } _ { \mathrm { B } } : \mathrm { v } _ { \mathrm { C } } = 1 : 1 : 1 , \omega _ { \mathrm { A } } : \omega _ { \mathrm { B } } : \omega _ { \mathrm { C } } = 2 : 3 : 6$
C. C. $\mathrm { v } _ { \mathrm { A } } : \mathrm { v } _ { \mathrm { B } } : \mathrm { v } _ { \mathrm { C } } = 1 : 1 : 1 , \omega _ { \mathrm { A } } : \omega _ { \mathrm { B } } : \omega _ { \mathrm { C } } = 1 : 2 : 3$
D. D. $\mathrm { v } _ { \mathrm { A } } : \mathrm { v } _ { \mathrm { B } } : \mathrm { v } _ { \mathrm { C } } = 3 : 2 : 1 , \omega _ { \mathrm { A } } : \omega _ { \mathrm { B } } : \omega _ { \mathrm { C } } = 1 : 1 : 1$

Answer

Answer: C

Solution: From the question, the device $A , B$ wheel is driven by belt, $A , B$ the points on the edge have the same linear velocity; $A , C$ wheel is driven by friction, $A , C$ the points on the edge have the same linear velocity. So the linear velocities of the three points are equal, then: $$ v _ { A } : v _ { B } : v _ { C } = 1 : 1 : 1 ; $$ According to the relationship between linear velocity and angular velocity: $v = \omega r$, we get : $$ \omega _ { A } : \omega _ { B } : \omega _ { C } = \frac { 1 } { R _ { A } } : \frac { 1 } { R _ { B } } : \frac { 1 } { R _ { C } } = 123 $$ A.$v A : v B : v C = 1 : 2 : 3 , \omega A : \omega B : \omega C = 3 : 2 : 1$ , which is not consistent with the conclusion, option A is wrong; B.$v A : v B : v C = 1 : 1 : 1 , \omega A : \omega B : \omega C = 2 : 3 : 6$ , which is inconsistent with the conclusion, option B is wrong; C. $v A : v B : v C = 1 : 1 : 1 , \omega A : \omega B : \omega C = 1 : 2 : 3$, which is consistent with the conclusion, option C is correct; D.$v A : v B : v C = 3 : 2 : 1 , \omega A : \omega B : \omega C = 1 : 1 : 1$ is inconsistent with the conclusion.

Question 22

Question 22: 22. At 23:55 Beijing time on March 26, 2025, the Xichang Satellite Launch Center Long March 3B carri...

22. At 23:55 Beijing time on March 26, 2025, the Xichang Satellite Launch Center Long March 3B carrier rocket ignited and took off, and the satellite Sky Link 2 04 entered the predetermined orbit smoothly, and the launch mission was a complete success. The launch can be simplified as the process shown in the figure, the satellite is first launched into the circular orbit $r$ with a radius of $I$ to do a uniform circular motion, and the satellite moves to the point of $A$ when it is derailed into the elliptical orbit II, and then moves to the apogee of the elliptical orbit II, which is the [$A$, and then moves to the apogee of the elliptical orbit II, which is the orbit of the satellite. INLINE_FORMULA_3]], the satellite changes orbit again and enters the circular orbit III with radius $2 r$ to do uniform circular motion. The following judgment is correct ![](/images/questions/phys-circular-motion/image-004.jpg)

A. A. The ratio of the period of the satellite orbiting in orbit I to orbit III is $1 : 2 \sqrt { 2 }$
B. B. To realize the entry into circular orbit III from the elliptical orbit $B$, the engine needs to be injected forward
C. C. The mechanical energy of the satellite in orbit I is greater than that in orbit III
D. D. The area swept per unit of time by the line to the center of the earth in orbit I and orbit II must be equal.

Answer

Answer: A

Solution: A. During the orbit of the satellite in orbit I and orbit III, according to Kepler's third law, there are $\frac { r ^ { 3 } } { T _ { 1 } ^ { 2 } } = \frac { ( 2 r ) ^ { 3 } } { T _ { 3 } ^ { 2 } }$ Solve for $T _ { 1 } : T _ { 3 } = 1 : 2 \sqrt { 2 }$ Therefore, A is correct; B. Orbit II is a low orbit relative to orbit III, and it needs to be accelerated at the tangent point in order to change orbit from low orbit to high orbit, i.e., the engine needs to be jetted backward in order to realize entering orbit III from the elliptical orbit $B$, so B is incorrect; C. Combined with the above, it can be seen that the satellite from orbit I to orbit III needs to be accelerated at $A$ and $B$, that is to say, the mechanical energy of the satellite in orbit I is less than that in orbit III, so C is wrong; D. According to Kepler's second law, in the same orbit, the area swept by the satellite and the geocentric line at equal times is equal, but in different orbits the area swept by the satellite and the geocentric line at equal times is not equal, that is, the area swept by the satellite and the geocentric line per unit of time is not equal on orbital 1 and orbital II, so it is wrong in D.

Question 23

Question 23: 23. On May 3, 2024, China launched the Chang'e 6 probe, initiating the first human mission to sample...

23. On May 3, 2024, China launched the Chang'e 6 probe, initiating the first human mission to sample and return to the back of the Moon. For this landing on the Moon, Chang'e 6 needs to go through the three re-orbiting processes shown in the figure (of which 1 is a circular orbit, and II and III are elliptical orbits), and then it will enter the landing process at an opportune time, and then enter the surface of the Moon. It is known that point $P$ is the cotangent point of the four orbits, point $Q$ is the far-moon point on orbit II, and the gravitational constant is $G$, then the following statements are correct ( ) ![](/images/questions/phys-circular-motion/image-005.jpg) Earth-Moon transfer orbit

A. A. Period of operation of Chang'e 6 while in orbit $T _ { \text {III } } < T _ { \text {II } } < T _ { \text {I } }$
B. B. If the orbit I approximate close to the lunar surface, known "Chang'e 6" in orbit I movement period, can be deduced that the density of the moon
C. C. Chang'e 6 passes the point $P$ in orbit II and that point in orbit I. Because of the different orbits, the acceleration is also different.
D. D. During the movement of Chang'e 6 from point $P$ to point $Q$ in orbit II, its mechanical energy gradually decreases due to the negative work done by the gravitational force.

Answer

Answer: B

Solution: A. According to Kepler's third law, Chang'e 6 has an orbital period of $T _ { \text {III } } > T _ { \text {II } } > T _ { \text {I } }$, so A is wrong; B. According to the centripetal force provided by gravity, when moving close to the surface of the Moon, the period of operation is $T _ { \text {III } } > T _ { \text {II } } > T _ { \text {I } }$, so A error $$ \frac { G M m } { R ^ { 2 } } = m \frac { 4 \pi ^ { 2 } } { T ^ { 2 } } R $$ According to the relationship between density and mass $$ \rho = \frac { M } { \frac { 4 } { 3 } \pi R ^ { 3 } } $$ the solution is obtained by association $$ \rho = \frac { 3 \pi } { G T ^ { 2 } } $$ Therefore, B is correct; C. According to $$ \frac { G M m } { r ^ { 2 } } = m a $$ C. According to $$ \frac { G M m } { r ^ { 2 } } = m a $$, it can be seen that the acceleration of Chang'e 6 is equal when it runs to point $P$ in Orbit I and Orbit II, so C is wrong; D. Chang'e-6 moves from point $P$ to point $Q$ in orbit II, only the gravitational force does work, and its mechanical energy remains unchanged, so D is wrong.

Question 24

Question 24: 24. We have learned that if an object in the force $F$ under the action of the force to move a dista...

24. We have learned that if an object in the force $F$ under the action of the force to move a distance in the direction of the force $l$, the force on the object to do the work $W = F l$, the unit of work is the joule (J). The joule is correctly expressed in terms of the base unit of the International System of Units: ( )

A. A. $\mathrm { kg } / \mathrm { s } ^ { 3 }$
B. B. $\mathrm { kg } \cdot \mathrm { m } ^ { 2 } / \mathrm { s } ^ { 3 }$
C. C. $\mathrm { kg } / \mathrm { s } ^ { 2 }$
D. D. $\mathrm { kg } \cdot \mathrm { m } ^ { 2 } / \mathrm { s } ^ { 2 }$

Answer

Answer: D

Solution: According to the definition of work $W = F l$ there are $$ 1 \mathrm {~J} = 1 \mathrm {~N} \cdot \mathrm {~m} = 1 \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s } ^ { 2 } \cdot \mathrm {~m} = 1 \mathrm {~kg} \cdot \mathrm {~m} ^ { 2 } / \mathrm { s } ^ { 2 } $$

Question 25

Question 25: 25. If the figure shows a clock with accurate timekeeping, the ratio of the angular velocity of the ...

25. If the figure shows a clock with accurate timekeeping, the ratio of the angular velocity of the minute hand to that of the hour hand is ( ).

A. A. $1 : 1$
B. B. $2 : 1$
C. C. $12 : 1$
D. D. $24 : 1$

Answer

Answer: C

Solution: The period of the minute hand is $$ T _ { 1 } = 1 \mathrm {~h} $$ The period of the hour hand is $$ T _ { 2 } = 12 \mathrm {~h} $$ According to $$ T = \frac { 2 \pi } { \omega } $$ It is known that the ratio of the angular velocity of the minute hand to that of the hour hand is $$ \frac { \omega _ { 1 } } { \omega _ { 2 } } = \frac { T _ { 2 } } { T _ { 1 } } = \frac { 12 } { 1 } $$

Question 26

Question 26: As shown in the figure, a weight $P$ is placed on a long wooden board $O A$, and the weight $P$ rema...

As shown in the figure, a weight $P$ is placed on a long wooden board $O A$, and the weight $P$ remains stationary with respect to the board as it is turned around the $O$ end of the board at a small angle. The weight $P$ remains stationary with respect to the board. The work done on the friction and elasticity of the board on the weight $P$ is: ![](/images/questions/phys-circular-motion/image-006.jpg)

A. A. Friction does no work on the weight
B. B. Friction does negative work on the weight
C. C. The elastic force does no work on the weight
D. D. The elastic force does negative work on the weight

Answer

Answer: A

Solution: The block is acted upon by gravity, support and friction, and in the process of turning a small angle, the displacement is in the upward direction, gravity does negative work on the block, and the elastic force does positive work on the object; the block is not sliding with respect to the board as it rises, so the friction force on the block does zero work on the block.

Question 27

Question 27: 27. The following statements are correct

27. The following statements are correct

A. A. Isaac Newton discovered the law of gravity and measured the value of the gravitational constant G.
B. B. Galileo's Dialogue of the Two New Sciences; Newton's Mathematical Principles of Natural Philosophy.
C. C. Newton invented the law of motion of falling bodies; Galileo discovered the law of inertia.
D. D. The Danish astronomer Tigu derived the laws of planetary motion by observing and studying the movements of the planets.

Answer

Answer: B

Solution: A. Newton discovered the law of gravity, and Cavendish measured the value of the gravitational constant $G$, so A is wrong; B. Galileo wrote Dialogue of Two New Sciences; Newton wrote Mathematical Principles of Natural Philosophy, so B is correct; C. Galileo invented the law of motion of falling bodies; Newton discovered the law of inertia, so C is wrong; D. Kepler through the planetary motion observation and study of the laws of planetary motion, so D error.

Question 28

Question 28: 28. The flight orbit of the Shenzhou VII spacecraft can be regarded as a near-Earth orbit, generally...

28. The flight orbit of the Shenzhou VII spacecraft can be regarded as a near-Earth orbit, generally over the Earth $300 \sim 700 \mathrm {~km}$, and the time for a one-week flight around the Earth is about 90 min, so that the number of times the astronauts in the Space Shuttle can see the sunrise and sunset in 24 h should be

A. A. 2
B. B. 4
C. C. 8
D. D. 16

Answer

Answer: D

Solution: The spacecraft sees the sunrise and sunset once every 90 min, the total flight time of the spacecraft $24 \mathrm {~h} = 1440 \mathrm {~min}$, so the total number of flight weeks $$ N = \frac { 1440 } { 90 } = 16 $$

Question 29

Question 29: 29. As shown in the figure, A, B two speedboats in the lake to do uniform circular motion, in the sa...

29. As shown in the figure, A, B two speedboats in the lake to do uniform circular motion, in the same time, they passed the distance ratio is $3 : 4$, the direction of motion to change the angle of the ratio is $3 : 2$, then they are ![](/images/questions/phys-circular-motion/image-007.jpg)

A. A. The ratio of the magnitude of the linear velocity is $4 : 3$
B. B. The ratio of angular velocity is $2 : 3$
C. C. The ratio of the radii of the circular motion is $1 : 2$
D. D. The ratio of the magnitude of centripetal acceleration is $1 : 2$

Answer

Answer: C

Solution: A. The ratio of the distances traveled by speedboats A and B in the same amount of time is $3 : 4$, and according to $v = \frac { \Delta s } { \Delta t }$, it follows that The ratio of their linear velocities is $3 : 4$, so A is wrong; B. The ratio of the angle of change in the direction of motion of two speedboats A and B in the same time is $3 : 2$, and the ratio of their angular velocities is $\omega = \frac { \Delta \theta } { \Delta t }$ according to $\omega = \frac { \Delta \theta } { \Delta t }$, so B is wrong; C. According to $v = \omega r$, the ratio of the radii of the two speedboats A and B in circular motion is $r _ { \mathrm { A } } : r _ { \mathrm { B } } = \frac { v _ { \mathrm { A } } } { \omega _ { \mathrm { A } } } : \frac { v _ { \mathrm { B } } } { \omega _ { \mathrm { B } } } = 1 : 2$, so C is correct; D. According to $a = \omega ^ { 2 } r = v \omega$ D. According to $a = \omega ^ { 2 } r = v \omega$, the ratio of the centripetal acceleration of the two speedboats is $a _ { \mathrm { A } } : a _ { \mathrm { B } } = v _ { \mathrm { A } } \omega _ { \mathrm { A } } : v _ { \mathrm { B } } \omega _ { \mathrm { B } } = 9 : 8$, so D is wrong.

Question 30

Question 30: 30. On March 31, 2019, China successfully sent the "Tiangong II 01 Star" into geosynchronous orbit. ...

30. On March 31, 2019, China successfully sent the "Tiangong II 01 Star" into geosynchronous orbit. Tiangong-2 is orbiting at a distance of 390 kilometers from the ground, and the "Sky Link II 01 Star" can provide relay services for data transmission between Tiangong-2 and the ground measurement and control station.

A. A. The Tien Chain 2-01 star can be located directly above Tiangong 2 at all times.
B. B. The Sky Link 2-01 satellite can continuously maintain direct communication with Tiangong-2.
C. C. In equal time, the two satellites have equal areas swept by their radii of operation
D. D. The acceleration of the Sky Link 2-01 star is greater than the centripetal acceleration of an object at the equator following the Earth's rotation.

Answer

Answer: D

Question 31

Question 31: 31. The students of an astronomy interest group, want to estimate the distance from the Sun to the E...

31. The students of an astronomy interest group, want to estimate the distance from the Sun to the Earth, had a serious discussion, if the known period of the Earth's rotation is $T$, the gravitational constant is $G$, through the search for information to get the ratio of the Sun's and Earth's mass is . The condition you think is also needed is

A. A. Earth's radius and Earth's rotation period
B. B. Period of revolution of the Moon around the Earth
C. C. Average density of the Earth and radius of the Sun
D. D. Earth's radius and the acceleration of gravity at the Earth's surface

Answer

Answer: D

Solution: For the Earth's gravity to provide centripetal force: $G \frac { M m } { r ^ { 2 } } = m r \frac { 4 \pi ^ { 2 } } { T ^ { 2 } }$ also by the gold substitution: $G m = g R ^ { 2 }$; and $\frac { M } { m } = N$ by the above three formula can be $r = \sqrt [ 3 ] { \frac { N T ^ { 2 } g R ^ { 2 } } { 4 \pi ^ { 2 } } }$, then the physical quantities are also needed is the Earth's radius R and the Earth's surface The acceleration of gravity at the Earth's surface, g, is $D$ is correct.

Question 32

Question 32: 32. In the development of physics, many physicists have made outstanding contributions, and the foll...

32. In the development of physics, many physicists have made outstanding contributions, and the following statements are correct

A. A. Galileo discovered the law of gravity.
B. B. Newton discovered the law of gravity.
C. C. Newton measured the gravitational constant in the lab.
D. D. Galileo measured the gravitational constant in his lab.

Answer

Answer: B

Solution: $\mathrm { A } , \mathrm {~B}$ Item: Newton discovered the law of gravity, so A is wrong and B is correct; C and D: Cavendish experimentally measured the gravitational constant, so C and D are wrong.

Question 33

Question 33: 33. Changes in the Earth's environment, we are all witnessed, perhaps one day in the future is reall...

33. Changes in the Earth's environment, we are all witnessed, perhaps one day in the future is really no longer suitable for human habitation, the movie "Wandering the Earth" in the choice is to take the Earth with the wandering, and in reality, some people choose to move to Mars. FORMULA_2]], the density of the Earth is $\rho$, then

A. A. Mars has a rotation period of $2 \sqrt { 2 } T$
B. B. The acceleration of gravity on the surface of Mars is approximately $\frac { 2 } { 5 } g$
C. C. The first cosmic velocity of Mars is approximately $\frac { 1 } { 5 } v$
D. D. Mars has a density of approximately $\frac { 5 } { 4 } \rho$

Answer

Answer: B

Solution: A. Based on $$ G \frac { M m } { r ^ { 2 } } = m \left( \frac { 2 \pi } { T ^ { \prime } } \right) ^ { 2 } r $$ Solve for $$ T ^ { \prime } = 2 \pi \sqrt { \frac { r ^ { 3 } } { G M } } $$ Since the radius of revolution of Mars is not known so its period of revolution cannot be determined, so A is wrong; B. According to $$ G \frac { M m } { r ^ { 2 } } = m g ^ { \prime } $$ the solution is $$ g ^ { \prime } = \frac { G M } { r ^ { 2 } } $$ Then the acceleration of gravity on the surface of Mars is approximately $$ g ^ { \prime } = \frac { \frac { 1 } { 10 } } { \frac { 1 } { 2 ^ { 2 } } } g = \frac { 2 } { 5 } g $$ So B is correct ; C. According to $$ G \frac { M m } { r ^ { 2 } } = m \frac { v ^ { \prime 2 } } { r } $$ the solution is $$ v ^ { \prime } = \sqrt { \frac { G M } { r } } $$ The first cosmic velocity of Mars is approximately $$ v ^ { \prime } = \sqrt { \frac { \frac { 1 } { 10 } } { \frac { 1 } { 2 } } } v = \sqrt { \frac { 1 } { 5 } } v $$ So C is wrong ; D. According to $$ M = \rho ^ { \prime } \frac { 4 } { 3 } \pi r ^ { 3 } $$ the solution is $$ \rho ^ { \prime } = \frac { 3 M } { 4 \pi r ^ { 3 } } $$ Then the density of Mars is about $$ \rho ^ { \prime } = \frac { \frac { 1 } { 10 } } { \left( \frac { 1 } { 2 } \right) ^ { 3 } } \rho = \frac { 4 } { 5 } \rho $$ So D is wrong ;

Question 34

Question 34: 34. As shown in Figure A is a pair of each other around the rotation of unequal mass of the double b...

34. As shown in Figure A is a pair of each other around the rotation of unequal mass of the double black hole system, the schematic diagram shown in Figure B, the double black holes $A , B$ in the gravitational force between each other under the action of its connecting line around the $O$ point to do the uniform circular motion, if the black holes $A$, $B$ do circular motion of the radius ratio is $1 : 2$, the following statement is true ([INLINE_FORMULA_4]) _2]], $B$ do circular motion of the radius ratio of $1 : 2$, the following statement is correct ( ) ![](/images/questions/phys-circular-motion/image-008.jpg) A ![](/images/questions/phys-circular-motion/image-009.jpg) B

A. A. The ratio of the angular velocities of black holes A and B in circular motion is $1 : 2$
B. B. The ratio of the magnitudes of the centripetal forces of black holes A and B in circular motion is $2 : 1$
C. C. The ratio of the linear velocities of black holes A and B in circular motion is $1 : 2$
D. D. The ratio of the masses of black holes A and B is $1 : 1$

Answer

Answer: C

Solution: A. Since the black holes A and B are in uniform circular motion around the same point on the line connecting the two, their periods and angular velocities are equal, and the ratio of their angular velocities is $1 : 1$, A is wrong; B. The centripetal force is provided by the gravitational force between each other, there are $\frac { G m _ { \mathrm { A } } m _ { \mathrm { B } } } { \left( r _ { \mathrm { A } } + r _ { \mathrm { B } } \right) ^ { 2 } } = m _ { \mathrm { A } } r _ { \mathrm { A } } \omega ^ { 2 } = m _ { \mathrm { B } } r _ { \mathrm { B } } \omega ^ { 2 }$ The ratio of the magnitudes of the centripetal forces is $1 : 1$, B is wrong; CD. From $m _ { \mathrm { A } } r _ { \mathrm { A } } \omega ^ { 2 } = m _ { \mathrm { B } } r _ { \mathrm { B } } \omega ^ { 2 } , r _ { \mathrm { A } } : r _ { \mathrm { B } } = 12$, we can see that the black hole $1 : 1$ is a black hole. The ratio of the masses of the black hole $\mathrm { A } , \mathrm {~B}$ is $2 : 1$, and the ratio of the linear velocities of the black hole $v = \omega r$ is $\mathrm { A } , \mathrm {~B}$ is $1 : 2$, which is $1 : 2$. C is correct and D is wrong.

Question 35

Question 35: 35. As shown in the figure, in the 1687 publication of the Mathematical Principles of Natural Philos...

35. As shown in the figure, in the 1687 publication of the Mathematical Principles of Natural Philosophy, Newton envisioned: the mass of $m$ objects from the mountains thrown horizontally, thrown out of the speed is large enough, the object will be around the Earth to do a uniform circular motion, as an artificial Earth satellite. If the Earth is regarded as a sphere of uniform mass distribution with a radius of $R$, and the height of the object from the ground is $h$, and the acceleration of gravity on the ground is $g$, then the following statements are correct ( ) ![](/images/questions/phys-circular-motion/image-010.jpg)

A. A. The acceleration of the object at the instant after it is thrown is $g$
B. B. The magnitude of the centripetal force on an object in uniform circular motion around the Earth is $\left( \frac { R } { R + h } \right) \mathrm { mg }$
C. C. The magnitude of the linear velocity of an object in uniform circular motion around the Earth is $R \sqrt { \frac { g } { R + h } }$
D. D. The period of uniform circular motion of an object around the Earth is $2 \pi \sqrt { \frac { R + h } { g } }$

Answer

Answer: C

Solution: A. Based on $$ \frac { G M m } { ( R + h ) ^ { 2 } } = m g ^ { \prime } $$ and also $$ \frac { G M m } { R ^ { 2 } } = m g $$ The conjunction, solves for $$ g ^ { \prime } = g \left( \frac { R } { R + h } \right) ^ { 2 } $$ Therefore, A is wrong; B. The magnitude of the centripetal force of an object in uniform circular motion around the Earth is $$ F _ { \mathrm { n } } = \frac { G M m } { ( R + h ) ^ { 2 } } $$ The solution is $$ F _ { \mathrm { n } } = m g \left( \frac { R } { R + h } \right) ^ { 2 } $$ Therefore, B is incorrect; C. According to $$ \frac { G M m } { ( R + h ) ^ { 2 } } = m \frac { v ^ { 2 } } { R + h } $$ the magnitude of the linear velocity of the object in uniform circular motion around the earth is solved to be $$ v = R \sqrt { \frac { g } { R + h } } $$ Therefore, C is correct; D. The period of the uniform circular motion of the object around the Earth is $$ T = \frac { 2 \pi ( R + h ) } { v } $$ The solution is $$ T = \frac { 2 \pi } { R } \sqrt { \frac { ( R + h ) ^ { 3 } } { g } } $$ Therefore, D is wrong.

Question 36

Question 36: 36. The following statements are correct

36. The following statements are correct

A. A. Uniform circular motion must be curvilinear motion with varying acceleration
B. B. Circular motion cannot be decomposed into two mutually perpendicular linear motions
C. C. Work is a scalar quantity, there are positive and negative values of work, and the positive and negative values of work indicate the magnitude of work
D. D. An object in curved motion must have a non-zero impulse of the combined external force at any $\Delta t$ time.

Answer

Answer: A

Solution: A. The acceleration of uniform circular motion points to the center of the circle and the direction keeps changing, which must be a curved motion with changing acceleration, so A is correct; B. Circular motion can be decomposed into two mutually perpendicular linear motion, so B is wrong; C. Work only size, no direction, is a scalar quantity, the positive and negative work said power or resistance to do work, so C error; D. If the object to do uniform circular motion of the object by the combined force is not zero, when $\Delta t = T$ time, the momentum change is zero, the root of the According to the momentum theorem $F t = \Delta p$, it can be seen that $\Delta t$ time the impulse of the combined external force is zero, so D is wrong.

Question 37

Question 37: 37. The asteroid "2012DA14" orbits the Sun on the outer side of the Earth and once passed by the Ear...

37. The asteroid "2012DA14" orbits the Sun on the outer side of the Earth and once passed by the Earth (at its closest approach) at a speed of about 28,000 kilometers per hour. Astronomers speculate that the next closest approach to the Earth by the asteroid 2012DA14 will be 33 years from now, given that the orbits of the asteroid and the Earth (both circular) are in the same plane and in the same direction, and given that the Earth orbits around the Sun with a period of one year, the asteroid orbits around the Sun with the period of one year, and the asteroid orbits around the Sun with the period of one year.

A. A. 1 year
B. B. 32 years
C. C. $\frac { 33 } { 32 }$ year
D. D. $\frac { 66 } { 65 }$ year

Answer

Answer: C

Solution: At the time of the asteroid's two closest distances from the Earth, the Earth makes one more revolution than the asteroid, the $$ \left( \frac { 2 \pi } { T _ { \text {地 } } } - \frac { 2 \pi } { T _ { \text {星 } } } \right) t = 2 \pi $$ Solve for $$ T _ { \text {星 } } = \frac { 33 } { 32 } \text { 年 } $$

Question 38

Question 38: 38. As shown in the figure, Mars and Jupiter orbit between an asteroid belt, assuming that the belt ...

38. As shown in the figure, Mars and Jupiter orbit between an asteroid belt, assuming that the belt of asteroids only by the Sun's gravity, and around the Sun to do uniform circular motion. The following statement is correct. ![](/images/questions/phys-circular-motion/image-011.jpg)

A. A. Every asteroid at equal distances from the Sun has an equal gravitational pull from the Sun
B. B. The acceleration of the inner planets in the asteroid belt is less than that of the outer planets
C. C. All planets in the asteroid belt rotate around the Sun with a linear velocity greater than that of the Earth's rotation.
D. D. All asteroids move around the Sun with a period greater than one year

Answer

Answer: D

Solution: A. Based on $$ F = G \frac { M m } { R ^ { 2 } } $$ Then the magnitude of the gravitational force on each asteroid at equal distance from the Sun cannot be determined because the mass of the asteroid is indeterminate, so A is incorrect; B. According to $$ a = \frac { G M } { R ^ { 2 } } $$ It can be seen that the larger the radius the smaller the acceleration, then the acceleration of the inner planets of the asteroid belt is greater than the acceleration of the outer planets, so B is wrong ; C. According to $$ v = \sqrt { \frac { G M } { R } } $$ According to $$ v = \sqrt { \frac { G M } { R } } $$, the larger the radius, the smaller the linear velocity, the asteroid belt planets around the Sun's linear velocity are smaller than the Earth's linear velocity, so C error; D. According to $$ T = 2 \pi \sqrt { \frac { R ^ { 3 } } { G M } } $$ It can be seen that the larger the radius the greater the period, then the period of each asteroid's motion around the sun is greater than one year, so D is correct;

Question 39

Question 39: 39. It is known that the radius of the Moon in circular motion around the Earth is equal to 60 times...

39. It is known that the radius of the Moon in circular motion around the Earth is equal to 60 times the radius of the Earth, then the ratio of the acceleration of the Moon in circular motion around the Earth to the acceleration of gravity on the surface of the Earth is

A. A. 60
B. B. 3600
C. C. $\frac { 1 } { 3600 }$
D. D. $\frac { 1 } { 60 }$

Answer

Answer: C

Solution: The gravitational force on an object on the surface of the earth is equal to its gravitational force, then there are $$ G \frac { M m } { R ^ { 2 } } = m g $$ The Moon moves in uniform circular motion around the Earth with a centripetal force provided by gravity, then we have $$ G \frac { M m ^ { \prime } } { ( 60 R ) ^ { 2 } } = m ^ { \prime } a _ { n } $$ Combining the above two equations can be obtained $$ a _ { n } : g = 1 : 3600 $$

Question 40

Question 40: 40. On January 20, 2021, China successfully launched the geostationary satellite "Tiantong 1" 03 fro...

40. On January 20, 2021, China successfully launched the geostationary satellite "Tiantong 1" 03 from the Xichang Satellite Launch Center. Lift-off, marking important progress in the construction of China's first satellite mobile communication system, the following statements about the satellite are correct High School Physics Assignment, October 29, 2025

A. A. Running rate between $7.9 \mathrm {~km} / \mathrm { s }$ and $11.2 \mathrm {~km} / \mathrm { s }$.
B. B. It can be directly over any point on the ground, but at a certain distance from the center of the ground
C. C. The angular velocity is not equal to the angular velocity of the Earth's rotation.
D. D. The centripetal acceleration is greater than the centripetal acceleration of the Moon around the Earth.

Answer

Answer: D

Solution: ABC.Because it is a geostationary satellite, it can only be in the equatorial plane, with an angular velocity equal to the angular velocity of the Earth's rotation, and a linear velocity less than $7.9 \mathrm {~km} / \mathrm { s } , \mathrm { ABC }$ Error; D. According to the formula $$ a = G \frac { M } { r ^ { 2 } } $$ Since the Moon's orbital radius is greater, the Moon's centripetal acceleration around the Earth is smaller.D is correct.

Circular Motion and Universal Gravitation

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