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AP Physics 1 : Universal Gravitation

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Example Question #654 : Ap Physics 1

Two asteroids, one with a mass of $7.12*10^{18}kg$ and the other with mass 5.33*10^8kg are $10*10^{10}m$ apart. What is the acceleration of the SMALLER asteroid?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

Possible Answers:

$2.53*10^{-5}\frac{m}{s^2}$

$9.41*10^{-7}\frac{m}{s^2}$

$3.79*10^5\frac{m}{s^2}$

$6.11*10^{-24}\frac{m}{s^2}$

$4.74*10^{-14}\frac{m}{s^2}$

Correct answer:

$4.74*10^{-14}\frac{m}{s^2}$

Explanation:

Given that Newton's second law is F=ma , we can find the acceleration by first determining the force.

To find the gravitational force, use Newton's law of universal gravitation:

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

We are given the constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.12*10^{18}kg)(5.33*10^{8}kg)}{(10*10^{10}m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.79*10^{27}kg}{10*10^{21}m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (3.79*10^5\frac{kg^2}{m^2})$

$F_G=2.53*10^{-5}N$

We now have values for both the mass and the force. Using the original equation, we can now solve for the acceleration.

F=ma

$2.53*10^{-5}N=(5.33*10^8kg)a$

$\frac{2.53*10^{-5}N}{5.33*10^8kg}{}=a$

$4.74*10^{-14}\frac{m}{s^2}=a$

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Example Question #655 : Ap Physics 1

Two asteroids, one with a mass of $7.12*10^{18}kg$ and the other with mass 5.33*10^8kg are $10*10^{10}m$ apart. What is the acceleration of the LARGER asteroid?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

Possible Answers:

$1.63*10^{-12}\frac{m}{s^2}$

$3.55*10^{-24}\frac{m}{s}$

$7.12*10^{-18}\frac{m}{s}$

$4.99*10^{-7}\frac{m}{s^2}$

$3.92*10^{-42}\frac{m}{s^2}$

Correct answer:

$3.55*10^{-24}\frac{m}{s}$

Explanation:

Given that Newton's second law is F=ma , we can find the acceleration by first determining the force.

To find the gravitational force, use Newton's law of universal gravitation:

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

We are given the constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.12*10^{18}kg)(5.33*10^{8}kg)}{(10*10^{10}m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.79*10^{27}kg}{10*10^{21}m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (3.79*10^5\frac{kg^2}{m^2})$

$F_G=2.53*10^{-5}N$

We now have values for both the mass and the force. Using the original equation, we can now solve for the acceleration.

F=ma

$2.53*10^{-5}N=(7.12*10^{18}kg)a$

$\frac{2.53*10^{-5}N}{7.12*10^{18}kg}{}=a$

$3.55*10^{-24}\frac{m}{s^2}=a$

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Example Question #656 : Ap Physics 1

Two satellites are a distance from each other in space. If one of the satellites has a mass of and the other has a mass of , which one will have the smaller acceleration?

Possible Answers:

Neither will have an acceleration

We need to know the value of the masses to solve

They will both have the same acceleration

Correct answer:

Explanation:

The formula for force and acceleration is Newton's 2nd law: F=ma . We know the mass, but first we need to find the force:

For this equation, use the law of universal gravitation:

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

We know from the first equation that a force is a mass times an acceleration. That means we can rearrange the equation for universal gravitation to look a bit more like that first equation:

$F=G\frac{m_1m_2}{r^2}$ can turn into: $F=m_2*\frac{G*m_1}{r^2}$ and $F=m_1*\frac{G*m_2}{r^2}$ , respectively.

We know that the forces will be equal, so set these two equations equal to each other:

$m_1*\frac{G*m_2}{r^2}=m_2*\frac{G*m_1}{r^2}$

The problem tells us that m_2=2m_1

$m_1*\frac{G*2m_1}{r^2}=2m_1*\frac{G*m_1}{r^2}$

Let's say that $\frac{G*m_1}{r^2}=a$ to simplify.

m_1*2a=2m_1*a

As you can see, the acceleration for m_1 is twice the acceleration for m_2 . Therefore the mass will have the smaller acceleration.

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Example Question #51 : Specific Forces

An asteroid with a mass of $10*10^{15}kg$ approaches the Earth. If they are 250,000,000m apart, what is the gravitational force exerted by the asteroid on the Earth?

$\small M_{Earth}=5.972*10^{24}kg$

$\small G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Possible Answers:

$6.7*10^{30}N$

$6.37*10^{13}N$

$1.59*10^{22}N$

$1.27*10^{14}N$

Correct answer:

$6.37*10^{13}N$

Explanation:

For this question, use the law of universal gravitation:

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

We are given the value of each mass, the distance (radius), and the gravitational constant. Using these values, we can solve for the force of gravity.

$F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(10*10^{15}kg)*(5.972*10^{24}kg)}{(250,000,000m)^2}$

$F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{5.972*10^{40}kg^2}{6.25*10^{16}m^2}$

$F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(9.5552*10^{23}\frac{kg^2}{m^2})$

$F=6.37*10^{13}N$

This force will apply to both objects in question. As it turns out, it does not matter which mass we're looking at; the force of gravity on each mass will be the same. This is supported by Newton's third law.

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Example Question #658 : Ap Physics 1

An asteroid with a mass of $10*10^{15}kg$ approaches the Earth. If they are 250,000,000m apart, what is the gravitational force exerted by the Earth on the asteroid?

$\small M_{Earth}=5.972*10^{24}kg$

$\small G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Possible Answers:

$1.59*10^{22}N$

$7.22*10^{14}N$

$6.37*10^{13}N$

$1.59*10^{40}N$

Correct answer:

$6.37*10^{13}N$

Explanation:

For this question, use the law of universal gravitation:

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

We are given the value of each mass, the distance (radius), and the gravitational constant. Using these values, we can solve for the force of gravity.

$F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(10*10^{15}kg)*(5.972*10^{24}kg)}{(250,000,000m)^2}$

$F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{5.972*10^{40}kg^2}{6.25*10^{16}m^2}$

$F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(9.5552*10^{23}\frac{kg^2}{m^2})$

$F=6.37*10^{13}N$

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Example Question #51 : Forces

An asteroid with a mass of $10*10^{15}kg$ approaches the Earth. If they are 250,000,000m apart, what is the asteroid's resultant acceleration?

$\small M_{Earth}=5.972*10^{24}kg$

$\small G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Possible Answers:

$0.00004\frac{m}{s^2}$

$0.00637\frac{m}{s^2}$

$2.133*10^{-8}\frac{m}{s^2}$

$156.99\frac{m}{s^2}$

$0.00323\frac{m}{s^2}$

Correct answer:

$0.00637\frac{m}{s^2}$

Explanation:

The relationship between force and acceleration is Newton's second law:

F=ma

We know the mass, but we will need to find the force. For this calculation, use the law of universal gravitation:

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

We are given the value of each mass, the distance (radius), and the gravitational constant. Using these values, we can solve for the force of gravity.

$F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(10*10^{15}kg)*(5.972*10^{24}kg)}{(250,000,000m)^2}$

$F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{5.972*10^{40}kg^2}{6.25*10^{16}m^2}$

$F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(9.5552*10^{23}\frac{kg^2}{m^2})$

$F=6.37*10^{13}N$

Now that we know the force, we can use this value with the mass of the asteroid to find its acceleration.

F=ma

$6.37*10^{13}N=(10*10^{15}kg)*a$

$\frac{6.37*10^{13}N}{10*10^{15}kg}{}=a$

$0.00637\frac{m}{s^2}=a$

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Example Question #660 : Ap Physics 1

An asteroid with a mass of $10*10^{15}kg$ approaches the Earth. If they are 250,000,000m away, what is the Earth's resultant acceleration?

$M_{Earth}=5.972*10^{24}kg$

$G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Possible Answers:

$6.37*10^{-3}\frac{m}{s^2}$

$0\frac{m}{s^2}$

$1.07*10^{-11}\frac{m}{s^2}$

$4.06*10^{-5}\frac{m}{s^2}$

$5.33\frac{m}{s^2}$

Correct answer:

$1.07*10^{-11}\frac{m}{s^2}$

Explanation:

The relationship between force and acceleration is Newton's second law:

F=ma

We know the mass, but we will need to find the force. For this calculation, use the law of universal gravitation:

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

We are given the value of each mass, the distance (radius), and the gravitational constant. Using these values, we can solve for the force of gravity.

$F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(10*10^{15}kg)*(5.972*10^{24}kg)}{(250,000,000m)^2}$

$F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{5.972*10^{40}kg^2}{6.25*10^{16}m^2}$

$F=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(9.5552*10^{23}\frac{kg^2}{m^2})$

$F=6.37*10^{13}N$

Now that we know the force, we can use this value with the mass of the Earth to find its acceleration.

F=ma

$6.37*10^{13}N=(5.972*10^{24}kg)*a$

$\frac{6.37*10^{13}N}{5.972*10^{24}kg}{}=a$

$1.07*10^{-11}\frac{m}{s^2}=a$

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Example Question #11 : Universal Gravitation

Two satellites are a distance from each other in space. If one of the satellites has a mass of and the other has a mass of , which one will have the greater acceleration?

Possible Answers:

They will have the same acceleration

We need to know the value of the masses to solve

The acceleration of each satellite will be zero

Correct answer:

Explanation:

The relationship between force and acceleration is Newton's second law:

F=ma

We know the masses, but first we need to find the forces in order to draw a conclusion about the satellites' accelerations. For this calculation, use the law of universal gravitation:

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

We can write this equation in terms of each object:

$F_1=m_1*\frac{G*m_2}{r^2}$

$F_2=m_2*\frac{G*m_1}{r^2}$

We know that the force applied to each object will be equal, so we can set these equations equal to each other.

$m_1*\frac{G*m_2}{r^2}=m_2*\frac{G*m_1}{r^2}$

We know that the second object is twice the mass of the first.

$m_1*\frac{G*2m_1}{r^2}=2m_1*\frac{G*m_1}{r^2}$

We can substitute for the acceleration to simplify.

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

m_1*2a=2m_1*a

The acceleration for m_1 is twice the acceleration for m_2 ; thus, the lighter mass will have the greater acceleration.

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Example Question #52 : Specific Forces

An astronaut lands on a new planet. She knows her own mass, , and the radius of the planet, . What other value must she know in order to find the mass of the new planet?

Possible Answers:

The planet's distance from Earth

The density of the planet

The force of gravity she exerts on the planet

Air pressure on the planet

The orbit of the planet

Correct answer:

The force of gravity she exerts on the planet

Explanation:

To find the relationship described in the question, we need to use the law of universal gravitation:

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

The question suggests that we know the radius and one of the masses, and asks us to solve for the other mass.

$F=G\frac{m_{astro}m_{planet}}{r^2}$

Since is a constant, if we know the mass of the astronaut and the radius of the planet, all we need is the force due to gravity to solve for the mass of the planet. According to Newton's third law, the force of the planet on the astronaut will be equal and opposite to the force of the astronaut on the planet; thus, knowing her force on the planet will allows us to solve the equation.

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Example Question #53 : Specific Forces

An astronaut lands on a planet with the same mass as Earth, but twice 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:

$\frac{1}{2}g$

$\frac{1}{4}g$

Correct answer:

$\frac{1}{4}g$

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 twice 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.

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

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

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

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

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:

$\frac{1}{4}g=a$

The acceleration due to gravity on this new planet will be one quarter of what it would be on Earth.

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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 #654 : Ap Physics 1

Example Question #655 : Ap Physics 1

Example Question #656 : Ap Physics 1

Example Question #51 : Specific Forces

Example Question #658 : Ap Physics 1

Example Question #51 : Forces

Example Question #660 : Ap Physics 1

Example Question #11 : Universal Gravitation

Example Question #52 : Specific Forces

Example Question #53 : Specific Forces

All AP Physics 1 Resources

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