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AP Physics 1 : Universal Gravitation

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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:

$\frac{g}{9}$

$\frac{g}{3}$

Correct answer:

Explanation:

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

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

For the new planet,

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

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

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

Possible Answers:

$3169\frac{m}{s}$

$7862\frac{m}{s}$

$4753\frac{m}{s}$

$942\frac{m}{s}$

$1286\frac{m}{s}$

Correct answer:

$3169\frac{m}{s}$

Explanation:

The equation to calculate the escape velocity from a planet is

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

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:

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

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Example Question #2 : Universal Gravitation

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

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

Possible Answers:

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

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

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

None of the other answers

Correct answer:

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

Explanation:

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

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

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:

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

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

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Example Question #1 : Universal Gravitation

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

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

Possible Answers:

$6.22*10^{-8}N$

$8.87*10^{-10}N$

$3.12*10^{-1}N$

$1.19*10^{-10}N$

$2.66*10^{-11}N$

Correct answer:

$1.19*10^{-10}N$

Explanation:

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

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

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

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

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

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

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Example Question #4 : Universal Gravitation

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

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

Possible Answers:

$1.25*10^{-19}N$

$2.5*10^{-10}N$

$1.1*10^{3}N$

$2.5*10^{-20}N$

$2.2*10^{8}N$

Correct answer:

$2.5*10^{-10}N$

Explanation:

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

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

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

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

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

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

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Example Question #5 : Universal Gravitation

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

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

Possible Answers:

2022.44kg

1602.98kg

1.989kg

2405.33kg

166.28kg

Correct answer:

166.28kg

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 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.

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

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

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

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

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

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

27648kg^2=M^2

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

166.28kg=M

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Example Question #6 : 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 force between them?

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

Possible Answers:

$5.98*10^{12}N$

$1.11*10^{30}N$

$1.49*10^{-32}N$

$2.99*10^{11}N$

$3.01*10^{-8}N$

Correct answer:

$2.99*10^{11}N$

Explanation:

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

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

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

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

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

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

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

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Example Question #651 : Ap Physics 1

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:

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

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

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

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

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

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 #652 : 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 gravitational force on the LARGER asteroid?

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

Possible Answers:

$3.55*10^{-6}N$

3.79*10^5N

$4.74*10^{-6}N$

$2.53*10^{-5}N$

$4.61*10^{-10}N$

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.

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

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

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

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

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 #653 : 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 gravitational force on the SMALLER asteroid?

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

Possible Answers:

$4.74*10^{-14}N$

$1.11*10^{12}N$

$2.53*10^{-5}N$

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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AP Physics 1 : Universal Gravitation

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #1 : Universal Gravitation

Example Question #1 : Universal Gravitation

Example Question #2 : Universal Gravitation

Example Question #1 : Universal Gravitation

Example Question #4 : Universal Gravitation

Example Question #5 : Universal Gravitation

Example Question #6 : Universal Gravitation

Example Question #651 : Ap Physics 1

Example Question #652 : Ap Physics 1

Example Question #653 : Ap Physics 1

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