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AP Physics 1 : Universal Gravitation

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Example Questions

Example Question #61 : Universal Gravitation

Radius of the moon: $1.737*10^{6}m$

Mass of moon: $7.35*10^{22}kg$

Jennifer is piloting her spaceship around the moon. How fast does she need to go to oribit the moon 200km above the surface.

Possible Answers:

$666\frac{m}{s}$

$2506\frac{m}{s}$

$1590\frac{m}{s}$

$790\frac{m}{s}$

Correct answer:

$1590\frac{m}{s}$

Explanation:

The radius of the orbit will be the radius of the moon plus the altitude of the orbit.

$r_{orbit}=altitude+r_{moon}$

Converting to and plugging in values:

$r_{orbit}=2.0*10^{5}+1.737*10^{6}m$

$r_{orbit}=1.94*10^{6}m$

Centripetal force will need to equal universal gravitational force

$m_{ship}\frac{v^2}{r}=G\frac{m_{ship}m_{moon}}{r^2}$

Solving for velocity

$v=\sqrt{\frac{Gm_{moon}}{r}}$

Plugging in values:

$v=\sqrt{\frac{6.674*10^{-11}*7.35*10^{22}}{1.94*10^6}}$

$v=1590\frac{m}{s}$

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Example Question #91 : Forces

For a planet of mass $2.34*10^{22}kg$ and diameter 12000km what is the force of gravity on an object of mass 12kg ?

Possible Answers:

114 N

.52 N

32 N

5.2N

10 N

Correct answer:

.52 N

Explanation:

To solve this we use the universal gravitation formula

$\frac{Gm_{1}m_{2}}{r^{2}}$

where G is the gravitational constant $6.67408*10^{-11} m^3 kg^{-1} s^{-2}$

and r is the radius of the planet $\frac{12000}{2}=6000km$

plugging everything in we get $F_{g}=.52 N$

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Example Question #62 : Universal Gravitation

In a fictional universe, two planets exist: one with a mass of 1500 kg and diameter of 1200m , the other with a mass of 7000kg and diameter of 1600m . The two planets' closest surfaces to one another are separated by a distance of 3000m . Assuming that the gravitational constant in this universe, , is $1000 \frac{N*m^{2}}{kg^{2}}$ , what is the gravitational force between the two planets?

Possible Answers:

542.4J

542.4N

54240 N

5424 N

5424 J

Correct answer:

542.4N

Explanation:

This question tests your understanding of gravitational force between two objects, and your ability to apply the formula for gravitational force.

The formula for gravitational force is as follows: $F_{g}= \frac{GM_{1}M_{2}}{r^{2}}$

In this example, we are asked to explicitly solve for the gravitational force between the two planets in this alternative universe. We are given the values for , $M_{1}$ , and $M_{2}$ . To solve for the gravitational force between the planets, we must first calculate the value of , or the distance between the centers of each planet. We are given the values of the diameters of each of the planets, and therefore, if we divide each value by , we have the values of the radii for each planet. To calculate the distance between the centers of the two planets, we must add the values of each planet's radius together, in addition to the distance between the closest surfaces of the planets. Thus, $r = \frac{1200m}{2} + \frac{1600m}{2} + 3000 m$ . After completing the arithmetic, you find that r= 4400 m .

Now, you have each of the relevant values to plug into the gravitational force formula. The calculation is shown below:

$F_{g}= \frac{GM_{1}M_{2}}{r^{2}}$

$F_{g}= \frac{(1000 \frac{N*m^{2}}{kg^{2}} * 1500 kg * 7000 kg)} {(4400 m)^{2}}$

$F_{g} = \frac{10500000000 N*m^{2}} {19360000 m^{2}}$

$F_{g} = 542.4 N$

Therefore, the gravitational force between the two planets is 542.4N .

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Example Question #63 : Universal Gravitation

In a fictional universe, a planet and a star exist: planet (mass 6000 kg , diameter 3000 m ), and star (mass 4000 kg , diameter 2000m ). The star remains in a geosynchronous orbit around the planet. The closest surfaces of the planet and the star are separated by a distance of 4000m . Assume that the gravitational constant, , in this universe is $1500 \frac{N*m^{2}}{kg^{2}}$ . What is the magnitude of the gravitational potential energy between these two bodies?

Possible Answers:

5,539 J

7,385,602 J

5,539 N

5,538,462 N

5,538,462 J

Correct answer:

5,538,462 J

Explanation:

This question tests your understanding of gravitational potential energy between two objects, and your ability to apply the formula for gravitational potential energy.

The formula for gravitational potential energy is as follows: $U_{g}= \frac{-GMm}{r}$

In this question, you are asked to solve for the magnitude of the gravitational potential energy between the planet and the star in this alternative universe. We are given the values for , , and . To solve for the gravitational potential energy between the planets, we must first calculate the value of , or the distance between the centers of the planet and the star. We are given the values of the diameters of the planet and the star, and therefore, if we divide each value by , we have the values of the radii for each. To calculate the distance between the centers of each, we must add the values of each radius together, in addition to the distance between the closest surfaces of the two bodies. Thus, $r = \frac{3000m}{2} + \frac{2000m}{2} + 4000 m$ . After completing the arithmetic, you find that r=6500m .

Now, you have each of the relevant values to plug into the gravitational potential energy formula. The calculation is shown below:

$U_{g}= \frac{-GMm}{r}$

Because we are asked to find the magnitude of the value, we can neglect the negative sign.

$U_{g}= \frac{1500\frac{N*m^{2}}{kg^{2}}*6000kg*4000kg}{6500m}$

$U_{g} = \frac{(36000000000 N*m^{2})} {6500 m}$

$U_{g} = 5,538,462 \frac{N}{m} = 5,538,462 J$

Therefore, the magnitude of the gravitational potential energy between the two bodies is 5,538,462 J .

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Example Question #61 : Universal Gravitation

Earth radius: 6370km

Earth mass: $5.97*10^{24}kg$

Gravity constant: $6.674*10^{-11}\frac{N*m^2}{kg^2}$

A space ship is in a perfectly circular orbit 450km above the earth. Determine its linear velocity.

Possible Answers:

$7.6*10^3 \frac{m}{s}$

$1.7*10^3 \frac{m}{s}$

$4.8*10^3 \frac{m}{s}$

$9.8*10^3 \frac{m}{s}$

$5.9*10^3\frac{m}{s}$

Correct answer:

$7.6*10^3 \frac{m}{s}$

Explanation:

In orbit, the magnitude of the centripetal force is in magnitude equal to the gravitational force:

$m_{ship}\frac{v^2}{r}=G\frac{m_{ship}m_{earth}}{r^2}$

Where is the linear velocity and is the distance from the center of the earth.

Solve for :

$v=\sqrt{\frac{G*m_{earth}}{r}}$

Plug in values, making sure to convert kilometers to meters to match the units in the answers:

$v=\sqrt{\frac{6.674*10^{-11}*5.97*10^{24}}{(6370+450)*10^3}}$

$v=7.6*10^3 \frac{m}{s}$

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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 #61 : Universal Gravitation

Example Question #91 : Forces

Example Question #62 : Universal Gravitation

Example Question #63 : Universal Gravitation

Example Question #61 : Universal Gravitation

All AP Physics 1 Resources

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