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AP Physics 1 : Universal Gravitation

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Example Questions

Example Question #1 : Universal Gravitation

A certain planet has three times the radius of Earth and nine times the mass. How does the acceleration of gravity at the surface of this planet (a_g) compare to the acceleration at the surface of Earth (g)?

Possible Answers:

$\displaystyle 9g$

$\displaystyle 3g$

$\displaystyle g$

$\frac{g}{3}$ $\displaystyle \frac{g}{3}$

$\frac{g}{9}$ $\displaystyle \frac{g}{9}$

Correct answer:

$\displaystyle g$

Explanation:

The acceleration of gravity is given by the equation $a_{g} = \frac{GM}{r^{2}}$ $\displaystyle a_{g} = \frac{GM}{r^{2}}$, where G is constant.

For Earth, $a_{g} = \frac{GM_{earth}}{r_{earth}^{2}} = g$ $\displaystyle a_{g} = \frac{GM_{earth}}{r_{earth}^{2}} = g$.

For the new planet,

$a_{g} = \frac{G(9M_{earth})}{\left ( 3r_{earth} \right )^2} = \frac{G(9M_{earth})}{9r_{earth{}}^2} = \frac{GM_{earth}}{r_{earth}^{2}}} = g$ $\displaystyle a_{g} = \frac{G(9M_{earth})}{\left ( 3r_{earth} \right )^2} = \frac{G(9M_{earth})}{9r_{earth{}}^2} = \frac{GM_{earth}}{r_{earth}^{2}}} = g$.

So, the acceleration is the same in both cases.

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Example Question #1 : Universal Gravitation

A new planet is discovered with mass $3.21\cdot 10^{23}kg$ $\displaystyle 3.21\cdot 10^{23}kg$ and with a diameter of $8.53\cdot 10^{3}km$ $\displaystyle 8.53\cdot 10^{3}km$. What is the lowest escape velocity required to escape this planet's gravitational pull?

$G=6.67\cdot10^{-11}\frac{N\cdot m^2}{kg^2}$ $\displaystyle G=6.67\cdot10^{-11}\frac{N\cdot m^2}{kg^2}$

Possible Answers:

$942\frac{m}{s}$ $\displaystyle 942\frac{m}{s}$

$3169\frac{m}{s}$ $\displaystyle 3169\frac{m}{s}$

$7862\frac{m}{s}$ $\displaystyle 7862\frac{m}{s}$

$1286\frac{m}{s}$ $\displaystyle 1286\frac{m}{s}$

$4753\frac{m}{s}$ $\displaystyle 4753\frac{m}{s}$

Correct answer:

$3169\frac{m}{s}$ $\displaystyle 3169\frac{m}{s}$

Explanation:

The equation to calculate the escape velocity from a planet is

$v_{e}=\sqrt{\frac{2GM_{p}}{R_{p}}}$ $\displaystyle v_{e}=\sqrt{\frac{2GM_{p}}{R_{p}}}$

The diameter of the planet is given and can be divided by two to determine the radius of the planet. By plugging in the given values, the escape velocity threshold can be determined:

$v_{e}=\sqrt{\frac{2(6.67\cdot10^{-11}\frac{N\cdot m^2}{kg^2})(3.21\cdot10^{23}kg)}{\frac{(8.53\cdot10^{6}m)}{}2}}=3169\frac{m}{s}$ $\displaystyle v_{e}=\sqrt{\frac{2(6.67\cdot10^{-11}\frac{N\cdot m^2}{kg^2})(3.21\cdot10^{23}kg)}{\frac{(8.53\cdot10^{6}m)}{}2}}=3169\frac{m}{s}$

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Example Question #2 : Universal Gravitation

There are two isolated stars orbiting each other. The first star has a mass of $2\times10^{12}kg$ $\displaystyle 2\times10^{12}kg$ and the second star has a mass of $4.5\times10^{14}kg$ $\displaystyle 4.5\times10^{14}kg$. If the stars are 2,000km away, what is the gravitational force felt by the first star?

$G=6.67\times10^{-11}\frac{m^3}{kg\cdot s}$ $\displaystyle G=6.67\times10^{-11}\frac{m^3}{kg\cdot s}$

Possible Answers:

None of the other answers

$\displaystyle 0N$

$1.50\mathrm{x}10^9N$ $\displaystyle 1.50\mathrm{x}10^9N$

$1.50\mathrm{x}10^4N$ $\displaystyle 1.50\mathrm{x}10^4N$

$1.50\mathrm{x}10^{14}N$ $\displaystyle 1.50\mathrm{x}10^{14}N$

Correct answer:

$1.50\mathrm{x}10^4N$ $\displaystyle 1.50\mathrm{x}10^4N$

Explanation:

We need to know Newton's law of universal gravitation to solve this problem:

$F = G \frac{m_1\cdot m_2}{r^2}$ $\displaystyle F = G \frac{m_1\cdot m_2}{r^2}$

It is important to note that both suns will feel the same gravitational force. However, since they have different masses, they will accelerate at different rates.

Plugging in the variables we have, we get:

$F = 6.67\mathrm{x}10^{-11} \frac{2\mathrm{x}10^{12}kg \cdot 4.5\mathrm{x}10^{14}kg}{(2\mathrm{x}10^6m)^2}$ $\displaystyle F = 6.67\mathrm{x}10^{-11} \frac{2\mathrm{x}10^{12}kg \cdot 4.5\mathrm{x}10^{14}kg}{(2\mathrm{x}10^6m)^2}$

$F = 1.50\mathrm{x}10^4N$ $\displaystyle F = 1.50\mathrm{x}10^4N$

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Example Question #1 : Universal Gravitation

Two satellites in space, each with a mass of 2000kg $\displaystyle 2000kg$, are 1500m $\displaystyle 1500m$ apart from each other. What is the force of gravity between them?

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

Possible Answers:

$6.22*10^{-8}N$ $\displaystyle 6.22*10^{-8}N$

$8.87*10^{-10}N$ $\displaystyle 8.87*10^{-10}N$

$3.12*10^{-1}N$ $\displaystyle 3.12*10^{-1}N$

$2.66*10^{-11}N$ $\displaystyle 2.66*10^{-11}N$

$1.19*10^{-10}N$ $\displaystyle 1.19*10^{-10}N$

Correct answer:

$1.19*10^{-10}N$ $\displaystyle 1.19*10^{-10}N$

Explanation:

To solve this problem, use Newton's law of universal gravitation:

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

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

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(2000kg) (2000kg)}{(1500m)^2})$ $\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(2000kg) (2000kg)}{(1500m)^2})$

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{4000000kg^2}{2250000m^2})$ $\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{4000000kg^2}{2250000m^2})$

$F_G=1.19*10^{-10}N$ $\displaystyle F_G=1.19*10^{-10}N$

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Example Question #3 : Universal Gravitation

Two satellites in space, each with a mass of 1723kg $\displaystyle 1723kg$, are 890m $\displaystyle 890m$ apart from each other. What is the force of gravity between them?

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

Possible Answers:

$1.25*10^{-19}N$ $\displaystyle 1.25*10^{-19}N$

$2.2*10^{8}N$ $\displaystyle 2.2*10^{8}N$

$2.5*10^{-20}N$ $\displaystyle 2.5*10^{-20}N$

$2.5*10^{-10}N$ $\displaystyle 2.5*10^{-10}N$

$1.1*10^{3}N$ $\displaystyle 1.1*10^{3}N$

Correct answer:

$2.5*10^{-10}N$ $\displaystyle 2.5*10^{-10}N$

Explanation:

To solve this problem, use Newton's law of universal gravitation:

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

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

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(1723kg )(1723kg)}{(890m)^2})$ $\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(1723kg )(1723kg)}{(890m)^2})$

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{2968729kg^2}{792100m^2})$ $\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{2968729kg^2}{792100m^2})$

$F_G=2.5*10^{-10}N$ $\displaystyle F_G=2.5*10^{-10}N$

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Example Question #43 : Specific Forces

Two satellites in space, each with equal mass, are 80m $\displaystyle 80m$ apart from each other. If the force of gravity between them is $2.88*10^{-10}N$ $\displaystyle 2.88*10^{-10}N$, what is the mass of each satellite?

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

Possible Answers:

$\displaystyle 2405.33kg$

$\displaystyle 2022.44kg$

1.989kg $\displaystyle 1.989kg$

$\displaystyle 1602.98kg$

$\displaystyle 166.28kg$

Correct answer:

$\displaystyle 166.28kg$

Explanation:

To solve this problem, use Newton's law of universal gravitation:

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

We are given the value of the force, the distance (radius), and the gravitational constant. We are also told that the masses of the two satellites are equal. Since the masses are equal, we can reduce the numerator of the law of gravitation to a single variable.

$F_G=G\frac{M^2}{r^2}$ $\displaystyle F_G=G\frac{M^2}{r^2}$

Now we can use our give values to solve for the mass.

$2.88*10^{-10}N=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{M^2}{(80m)^2})$ $\displaystyle 2.88*10^{-10}N=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{M^2}{(80m)^2})$

$\frac{2.88*10^{-10}N}{(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})}=\frac{M^2}{6400m^2}$ $\displaystyle \frac{2.88*10^{-10}N}{(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})}=\frac{M^2}{6400m^2}$

$4.32\frac{kg^2}{m^2}=\frac{M^2}{6400m^2}$ $\displaystyle 4.32\frac{kg^2}{m^2}=\frac{M^2}{6400m^2}$

$6400m^2* 4.32\frac{kg^2}{m^2}=M^2$ $\displaystyle 6400m^2* 4.32\frac{kg^2}{m^2}=M^2$

$\displaystyle 27648kg^2=M^2$

$\sqrt{27648kg^2}=\sqrt{M^2}$ $\displaystyle \sqrt{27648kg^2}=\sqrt{M^2}$

$\displaystyle 166.28kg=M$

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Example Question #44 : Specific Forces

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

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

Possible Answers:

$2.99*10^{11}N$ $\displaystyle 2.99*10^{11}N$

$1.49*10^{-32}N$ $\displaystyle 1.49*10^{-32}N$

$1.11*10^{30}N$ $\displaystyle 1.11*10^{30}N$

$3.01*10^{-8}N$ $\displaystyle 3.01*10^{-8}N$

$5.98*10^{12}N$ $\displaystyle 5.98*10^{12}N$

Correct answer:

$2.99*10^{11}N$ $\displaystyle 2.99*10^{11}N$

Explanation:

To solve this problem, use Newton's law of universal gravitation:

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

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

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

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

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

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

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Example Question #1 : Understanding Universal Gravitation

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Given that F=ma $\displaystyle F=ma$, 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:

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

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

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

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

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

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

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

F=ma $\displaystyle F=ma$

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

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

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

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Example Question #4 : Universal Gravitation

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

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

Possible Answers:

$4.61*10^{-10}N$ $\displaystyle 4.61*10^{-10}N$

$4.74*10^{-6}N$ $\displaystyle 4.74*10^{-6}N$

$2.53*10^{-5}N$ $\displaystyle 2.53*10^{-5}N$

$3.55*10^{-6}N$ $\displaystyle 3.55*10^{-6}N$

$\displaystyle 3.79*10^5N$

Correct answer:

$2.53*10^{-5}N$ $\displaystyle 2.53*10^{-5}N$

Explanation:

To solve this problem, use Newton's law of universal gravitation:

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

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

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

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

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

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

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 #45 : Specific Forces

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

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

Possible Answers:

$2.53*10^{-5}N$ $\displaystyle 2.53*10^{-5}N$

$1.11*10^{12}N$ $\displaystyle 1.11*10^{12}N$

$3.55*10^{-24}N$ $\displaystyle 3.55*10^{-24}N$

$4.74*10^{-14}N$ $\displaystyle 4.74*10^{-14}N$

$\displaystyle 3.79*10^5N$

Correct answer:

$2.53*10^{-5}N$ $\displaystyle 2.53*10^{-5}N$

Explanation:

To solve this problem, use Newton's law of universal gravitation:

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

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

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

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

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

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AP Physics 1 : Universal Gravitation

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #1 : Universal Gravitation

Example Question #1 : Universal Gravitation

Example Question #2 : Universal Gravitation

Example Question #1 : Universal Gravitation

Example Question #3 : Universal Gravitation

Example Question #43 : Specific Forces

Example Question #44 : Specific Forces

Example Question #1 : Understanding Universal Gravitation

Example Question #4 : Universal Gravitation

Example Question #45 : Specific Forces

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