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AP Physics 1 : Universal Gravitation

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Example Question #21 : Universal Gravitation

An astronaut lands on a planet with twelve times the mass of Earth and the same radius. What will be the acceleration due to gravity on this planet, in terms of the acceleration due to gravity on Earth?

Possible Answers:

12g

$\frac{1}{12}g$

144g

$\frac{1}{144}g$

$g\sqrt{12}$

Correct answer:

12g

Explanation:

For this comparison, we can use the law of universal gravitation and Newton's second law:

$F=G\frac{m_1m_2}{r^2}$

F=ma

We know that the force due to gravity on Earth is equal to . We can use this to set the two force equations equal to one another.

$G\frac{m*m_{E}}{r^2}=mg$

Notice that the mass cancels out from both sides.

$G\frac{m_{E}}{r^2}=g$

This equation sets up the value of acceleration due to gravity on Earth.

The new planet has a mass equal to twelve times that of Earth. That means it has a mass of 12m_E . It has the same radius as Earth, . Using these variables, we can set up an equation for the acceleration due to gravity on the new planet.

$G\frac{12m_{E}}{r^2}=a$

$12(G\frac{m_{E}}{r^2})=a$

We had previously solved for the gravity on Earth:

$G\frac{m_{E}}{r^2}=g$

We can substitute this into the new acceleration equation:

12g=a

The acceleration due to gravity on this new planet will be twelve times what it would be on Earth.

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Example Question #21 : Universal Gravitation

An astronaut lands on a planet with three times the mass of Earth, and the same radius. What will be the acceleration due to gravity on this planet, in terms of the acceleration due to gravity on Earth?

Possible Answers:

$\frac{1}9{}g$

$\frac{1}{3}g$

Correct answer:

Explanation:

For this comparison, we can use the law of universal gravitation and Newton's second law:

$F=G\frac{m_1m_2}{r^2}$

F=ma

We know that the force due to gravity on Earth is equal to . We can use this to set the two force equations equal to one another.

$G\frac{m*m_{E}}{r^2}=mg$

Notice that the mass cancels out from both sides.

$G\frac{m_{E}}{r^2}=g$

This equation sets up the value of acceleration due to gravity on Earth.

The new planet has a mass equal to three times that of Earth. That means it has a mass of 3m_E . It has the same radius as Earth, . Using these variables, we can set up an equation for the acceleration due to gravity on the new planet.

$G\frac{3m_{E}}{r^2}=a$

$3*(G\frac{m_{E}}{r^2})=a$

We had previously solved for the gravity on Earth:

$G\frac{m_{E}}{r^2}=g$

We can substitute this into the new acceleration equation:

3g=a

The acceleration due to gravity on this new planet will be three times what it would be on Earth.

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Example Question #21 : Universal Gravitation

An astronaut lands on a planet with the same mass as Earth, but six times the radius. What will be the acceleration due to gravity on this planet, in terms of the acceleration due to gravity on Earth?

Possible Answers:

$g\sqrt6$

36g

$\frac{1}{36}g$

$\frac{1}{6}g$

Correct answer:

$\frac{1}{36}g$

Explanation:

For this comparison, we can use the law of universal gravitation and Newton's second law:

$F=G\frac{m_1m_2}{r^2}$

F=ma

We know that the force due to gravity on Earth is equal to . We can use this to set the two force equations equal to one another.

$G\frac{m*m_{E}}{r^2}=mg$

Notice that the mass cancels out from both sides.

$G\frac{m_{E}}{r^2}=g$

This equation sets up the value of acceleration due to gravity on Earth.

The new planet has a radius equal to six times that of Earth. That means it has a radius of . It has the same mass as Earth, m_E . Using these variables, we can set up an equation for the acceleration due to gravity on the new planet.

$G\frac{m_{E}}{(6r)^2}=a$

Expand this equation to compare it to the acceleration of gravity on Earth.

$G\frac{m_{E}}{36r^2}=a$

$\frac{1}{36}(G\frac{m_{E}}{r^2})=a$

We had previously solved for the gravity on Earth:

$G\frac{m_{E}}{r^2}=g$

We can substitute this into the new acceleration equation:

$\frac{1}{36}g=a$

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Example Question #61 : Specific Forces

An astronaut lands on a planet with six times the mass of Earth, and four times the radius. What will be the acceleration due to gravity on this planet, in terms of the acceleration due to gravity on Earth?

Possible Answers:

$\frac{2}{3}g$

$\frac{3}{8}g$

$\frac{3}{2}g$

$\frac{8}{3}g$

Correct answer:

$\frac{3}{8}g$

Explanation:

For this comparison, we can use the law of universal gravitation and Newton's second law:

$F=G\frac{m_1m_2}{r^2}$

F=ma

We know that the force due to gravity on Earth is equal to . We can use this to set the two force equations equal to one another.

$G\frac{m*m_{E}}{r^2}=mg$

Notice that the mass cancels out from both sides.

$G\frac{m_{E}}{r^2}=g$

This equation sets up the value of acceleration due to gravity on Earth.

The new planet has a mass equal to six times that of Earth. That means it has a mass of 6m_E . It also has four times the radius of Earth, . Using these variables, we can set up an equation for the acceleration due to gravity on the new planet.

$G\frac{6m_{E}}{(4r)^2}=a$

Expand this equation in order to combine the non-variable terms.

$G\frac{6m_{E}}{16r^2}=a$

$\frac{6}{16}(G\frac{m_{E}}{r^2})=a$

$\frac{3}{8}(G\frac{m_{E}}{r^2})=a$

We had previously solved for the gravity on Earth:

$G\frac{m_{E}}{r^2}=g$

We can substitute this into the new acceleration equation:

$\frac{3}{8}g=a$

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Example Question #61 : Specific Forces

An astronaut lands on a planet with twice the mass of Earth, and half of the radius. What will be the acceleration due to gravity on this planet, in terms of the acceleration due to gravity on Earth?

Possible Answers:

$\frac{1}{2}g$

Correct answer:

Explanation:

For this comparison, we can use the law of universal gravitation and Newton's second law:

$F=G\frac{m_1m_2}{r^2}$

F=ma

We know that the force due to gravity on Earth is equal to . We can use this to set the two force equations equal to one another.

$G\frac{m*m_{E}}{r^2}=mg$

Notice that the mass cancels out from both sides.

$G\frac{m_{E}}{r^2}=g$

This equation sets up the value of acceleration due to gravity on Earth.

The new planet has a mass equal to twice that of Earth. That means it has a mass of 2m_E . It also has half the radius of Earth, $\small \frac{1}{2}r$ . Using these variables, we can set up an equation for the acceleration due to gravity on the new planet.

$G\frac{2m_{E}}{(\frac{1}{2}r)^2}=a$

Expand this equation in order to combine the non-variable terms.

$G\frac{2m_{E}}{\frac{1}{4}r^2}=a$

$\frac{2}{\frac{1}{4}}(G\frac{m_{E}}{r^2})=a$

$8(G\frac{m_{E}}{r^2})=a$

We had previously solved for the gravity on Earth:

$G\frac{m_{E}}{r^2}=g$

We can substitute this into the new acceleration equation:

8g=a

The acceleration due to gravity on this new planet will be eight times what it would be on Earth.

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Example Question #21 : Universal Gravitation

An astronaut lands on a new planet. She observes that the force due to gravity acting upon her is half what it is on Earth. If the new planet has the same radius as Earth, what is the mass of this planet in terms of the mass of Earth?

Possible Answers:

$\frac{1}{4}m_E$

m_E

2m_E

4m_E

$\frac{1}{2}m_E$

Correct answer:

$\frac{1}{2}m_E$

Explanation:

For this comparison we can use the law of universal gravitation:

$F=G\frac{m_1m_2}{r^2}$

The astronaut notices that the force due to gravity on this planet is half what it is on Earth. We can thus relate the force equations for the two planets.

$F_{p}=\frac{1}{2}F_G$

$\frac{1}{2}(G\frac{m_1m_E}{r^2})=G\frac{m_1m_p}{r^2}$

m_1 refers to any arbitrary mass on the surface of the planet, and will be constant. The radii of the two planets are equal, and the gravitational constant is also equal. These three variables can cancel out from both sides of the equation.

$\frac{1}{2}m_E(G\frac{m_1}{r^2})=m_p(G\frac{m_1}{r^2})$

$\frac{1}{2}m_E=m_p$

This relation shows us that the mass of the new planet will be half that of Earth.

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Example Question #661 : Ap Physics 1

An astronaut lands on a new planet. She observes that the force due to gravity acting upon her is twice what it is on Earth. If the new planet has the same radius as Earth, what is the mass of this planet in terms of the mass of Earth?

Possible Answers:

4m_E

8m_E

$\frac{1}{4}m_E$

$2m_{E}$

$\frac{1}{2}m_E$

Correct answer:

$2m_{E}$

Explanation:

For this comparison we can use the law of universal gravitation:

$F=G\frac{m_1m_2}{r^2}$

The astronaut notices that the force due to gravity on this planet is twice what it is on Earth. We can thus relate the force equations for the two planets.

$F_{p}=2F_G$

$2(G\frac{m_1m_E}{r^2})=G\frac{m_1m_p}{r^2}$

$2m_E(G\frac{m_1}{r^2})=m_p(G\frac{m_1}{r^2})$

2m_E=m_p

This relation shows us that the mass of the new planet will be twice that of Earth.

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Example Question #671 : Ap Physics 1

Two asteroids exert a gravitational force on one another. By what factor would this force change if one asteroid doubles in mass, the other asteroid triples in mass, and the distance between them is quadrupled?

Possible Answers:

$\frac{1}{2}$

$\frac{8}{9}$

$\frac{3}{8}$

$\frac{8}{3}$

Correct answer:

$\frac{3}{8}$

Explanation:

The equation for the force of gravity between two objects is:

$F=\frac{Gm_1m_2}{r^2}$

Using this equation, we can select arbitrary values for our original masses and distance. This will make it easier to solve when these values change.

$m_1=1kg;\ m_2=1kg;\ r=1m$

$F_1=\frac{G(1kg)(1kg)}{(1m)^2}=G$

$\small G$ is the gravitational constant. Now that we have a term for the initial force of gravity, we can use the changes from the question to find how the force changes.

$m_{1a}=2m_1=2kg;\ m_{2a}=3m_2=3kg;\ r_a=4r=4m$

$F_2=\frac{G(2kg)(3kg)}{(4m)^2}$

$F_2=\frac{6kg^2G}{16m^2}$

$F_2=\frac{3}{8}G$

We can use our first calculation to see the how the force has changed.

$F_1=G\ \text{and}\ F_2=\frac{3}{8}G$

$F_2=\frac{3}{8}F_1$

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Example Question #66 : Specific Forces

A 2500kg satellite orbits 3.8*10^8m above the Earth. What is the gravitational force of the Earth on the satellite?

$m_E=5.97*10^{24}kg$

r_E=6.37*10^5m

$G=6.67*10^{-11}\frac{m^3}{kg*s}$

Possible Answers:

$1.56*10^{12}N$

$1.03*10^{11}N$

$6.63*10^{-3}N$

1.03N

6.87N

Correct answer:

6.87N

Explanation:

To solve this problem, use the law of universal gravitation.

$F_G=G\frac{m_1m_2}{r^2}$

Remember that is the distance between the centers of the two objects. That means it will be equal to the radius of the earth PLUS the orbiting distance.

Use the given values for the masses of the objects and distance to solve for the force of gravity.

$F_G=G\frac{m_sm_E}{(r_E+h)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s})*\frac{(2500kg)(5.97*10^{24}kg)}{(6.37*10^5m+3.8*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.49*10^{28}kg^2}{1.45*10^{17}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(1.03*10^{11}\frac{kg^2}{m^2})$

F_G=6.87N

This force will be the same for both objects. The Earth will exert the same force on the satellite as the satellite exerts on the Earth. This makes sense, given Newton's third law.

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Example Question #67 : Specific Forces

A 2500kg satellite orbits 3.8*10^8m above the Earth. What is the gravitational force of the satellite on the Earth?

$m_E=5.97*10^{24}kg$

r_E=6.37*10^5m

$G=6.67*10^{-11}\frac{m^3}{kg*s}$

Possible Answers:

68.7N

6.87N

$1.03*10^{11}N$

0.15N

$1.5*10^{12}N$

Correct answer:

6.87N

Explanation:

To solve this problem, use the law of universal gravitation.

$F_G=G\frac{m_1m_2}{r^2}$

Remember that is the distance between the centers of the two objects. That means it will be equal to the radius of the earth PLUS the orbiting distance.

Use the given values for the masses of the objects and distance to solve for the force of gravity.

$F_G=G\frac{m_sm_E}{(r_E+h)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s})*\frac{(2500kg)(5.97*10^{24}kg)}{(6.37*10^5m+3.8*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.49*10^{28}kg^2}{1.45*10^{17}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(1.03*10^{11}\frac{kg^2}{m^2})$

F_G=6.87N

This force will be the same for both objects. The Earth will exert the same force on the satellite as the satellite exerts on the Earth. This makes sense, given Newton's third law.

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AP Physics 1 : Universal Gravitation

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #21 : Universal Gravitation

Example Question #21 : Universal Gravitation

Example Question #21 : Universal Gravitation

Example Question #61 : Specific Forces

Example Question #61 : Specific Forces

Example Question #21 : Universal Gravitation

Example Question #661 : Ap Physics 1

Example Question #671 : Ap Physics 1

Example Question #66 : Specific Forces

Example Question #67 : Specific Forces

All AP Physics 1 Resources

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