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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #3 : Universal Gravitation

Two asteroids in space are in close proximity to each other. Each has a mass of $6.69*10^{15}kg$ . If they are 100,000m apart, what is the gravitational acceleration that they experience?

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

Possible Answers:

$5.12*10^{4}\frac{m}{s^2}$

$3.89*10^{10}\frac{m}{s^2}$

$2.99*10^{11}\frac{m}{s^2}$

$6.69*10^{15}\frac{m}{s^2}$

$4.47*10^{-5}\frac{m}{s^2}$

Correct answer:

$4.47*10^{-5}\frac{m}{s^2}$

Explanation:

Given that F=ma , we already know the mass, but we need to find the force in order to solve for the acceleration.

To solve this problem, 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 satellite masses and distance (radius). Using these values we can solve for the force.

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(\frac{(6.69*10^{15}kg)(6.69*10^{15}kg)}{(100,000m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{4.48*10^{31}kg^2}{10*10^9m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(4.48*10^{21}\frac{kg^2}{m^2})$

$F_G=2.99*10^{11}N$

Now we have values for both the mass and the force, allowing us to solve for the acceleration.

F=ma

$2.99*10^{11}N=(6.69*10^{15}kg)a$

$\frac{2.99*10^{11}N}{6.69*10^{15}kg}{}=a$

$4.47*10^{-5}\frac{m}{s^2}=a$

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Example Question #2 : Universal Gravitation

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 gravitational force on the LARGER asteroid?

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

Possible Answers:

$3.55*10^{-6}N$

$4.74*10^{-6}N$

$2.53*10^{-5}N$

$4.61*10^{-10}N$

3.79*10^5N

Correct answer:

$2.53*10^{-5}N$

Explanation:

To solve this problem, 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.

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

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

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

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

It actually doesn't matter which asteroid we're looking at; the gravitational force will be the same. This makes sense because Newton's 3rd law states that the force one asteroid exerts on the other is equal in magnitude, but opposite in direction, to the force the other asteroid exerts on it.

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Example Question #3 : Universal Gravitation

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 gravitational force on the SMALLER asteroid?

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

Possible Answers:

$1.11*10^{12}N$

$4.74*10^{-14}N$

$2.53*10^{-5}N$

3.79*10^5N

$3.55*10^{-24}N$

Correct answer:

$2.53*10^{-5}N$

Explanation:

To solve this problem, 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.

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

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

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

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

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Example Question #11 : Universal Gravitation

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:

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

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

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

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

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

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 #2 : Understanding Universal Gravitation

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:

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

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

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

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

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

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 #12 : 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 smaller acceleration?

Possible Answers:

We need to know the value of the masses to solve

They will both have the same acceleration

Neither will have an 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 #3 : Understanding Universal Gravitation

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:

$6.37*10^{13}N$

$1.27*10^{14}N$

$1.59*10^{22}N$

$6.7*10^{30}N$

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 #11 : Understanding Universal Gravitation

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:

$1.59*10^{22}N$

$6.37*10^{13}N$

$7.22*10^{14}N$

$1.59*10^{40}N$

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 #13 : Universal Gravitation

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:

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

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

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

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

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

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 #14 : Universal Gravitation

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:

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

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

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

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

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

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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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #3 : Universal Gravitation

Example Question #2 : Universal Gravitation

Example Question #3 : Universal Gravitation

Example Question #11 : Universal Gravitation

Example Question #2 : Understanding Universal Gravitation

Example Question #12 : Universal Gravitation

Example Question #3 : Understanding Universal Gravitation

Example Question #11 : Understanding Universal Gravitation

Example Question #13 : Universal Gravitation

Example Question #14 : Universal Gravitation

All AP Physics 1 Resources

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