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AP Physics 1 : Universal Gravitation

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Example Questions

Example Question #641 : Newtonian Mechanics

How high would you have to be above the Earth’s surface to experience an acceleration of $\frac{g}{9}$ ? (Be careful!)

Possible Answers:

3R_E

$\sqrt{3}R_E$

2R_E

$\sqrt{2}R_E$

R_E

Correct answer:

2R_E

Explanation:

This problem can be solved by looking at the formula for the gravitational constant, .

$g=G\frac{M_E}{R_E^2}$

We are looking for the altitude above the Earth where the gravitation field is $\frac{1}{9}$ in strength, so we proceed as follows

$\frac{g}{9}=\frac{GM_E}{\left(dR_E \right )^2}\Rightarrow d=3R_E\Rightarrow r=d-R_E=2R_E$

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Example Question #681 : Ap Physics 1

Two gaseous planets $P_{1}$ and $P_{2}$ orbiting a distant star have equal masses, but $P_{1}$ has 6 times the radius of $P_{2}$ . What is the relationship between the acceleration of gravity at the surface of these two planets?

Possible Answers:

$g_{2}=\frac{g_{1}}{36}$

$g_{1}=\frac{g_{2}}{6}$

$g_{1}=\frac{g_{2}}{36}$

None of these.

$g_{2}=\frac{g_{1}}{12}$

Correct answer:

$g_{1}=\frac{g_{2}}{36}$

Explanation:

Use Newton's law of gravitation to make sense of this problem:

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

As you can see the force of gravity is inversely proportional to the square of the radius. Since $P_{1}$ has a radius 6 times larger than $P_{2}$ , we state that $g_{1}=\frac{g_{2}}{36}$ because the acceleration of the gravity of planet two divided by 36 would equal the smaller gravitational acceleration of planet one.

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Example Question #641 : Newtonian Mechanics

Moon radius: 1,737.5 km

Moon mass: $7.34*10^{22} kg$

$G=6.674*10^{-11} \frac{N*m^2}{kg^2}$

Calculate the gravity constant for objects on the surface of the moon.

Possible Answers:

$g_{moon}=1.62\frac{m}{s^2}$

$g_{moon}=11.22\frac{m}{s^2}$

$g_{moon}=4.55\frac{m}{s^2}$

$g_{moon}=2.44\frac{m}{s^2}$

$g_{moon}=6.13\frac{m}{s^2}$

Correct answer:

$g_{moon}=1.62\frac{m}{s^2}$

Explanation:

Universal Gravitation law:

$F_{grav}=-G\frac{m_1m_2}{r^2}$

Assume m_1 is the Moon, and m_2 is the object on the surface, then:

$-G\frac{m_1}{r^2}$

Will be analogous to in $F_{grav}=mg$ , which will be a negative number since it is pointing down.

Plug in values:

$-6.674*10^{-11}\frac{7.34*10^{22}}{(1,740*10^3)^2}=g_{moon}$

$g_{moon}=1.62\frac{m}{s^2}$

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Example Question #642 : Newtonian Mechanics

The center of the moon is 3.84*10^8m from the earth.

The earth has a mass of $6.0*10^{24}kg$ .

Estimate lunar velocity.

Possible Answers:

None of these

$2.14*10^6\frac{m}{s}$

$.84*10^6\frac{m}{s}$

$1.04*10^6\frac{m}{s}$

$1.51*10^6\frac{m}{s}$

Correct answer:

$1.04*10^6\frac{m}{s}$

Explanation:

In an orbit, the centripital force will be equal to the gravitational force:

$m_m\frac{v^2}{r}=G\frac{m_e*m_m}{r^2}$

Where:

is the distance to the center of the earth from the center of the moon.

is the linear velocity of the moon

m_m is the mass of the moon

m_e is the mass of the earth

is the universal gravity constant, $6.674*10^{-11}N*\frac{m^2}{kg^2}$

Simplifying and plugging in values:

${v^2}=6.67*10^{-11}\frac{7.35*10^{22}}{1.75*10^6}$

Solving for :

$v=1.04*10^6\frac{m}{s}$

*note how the mass of the moon was irrelevant

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Example Question #643 : Newtonian Mechanics

Distance from earth to the moon: 384000 km

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

Universal gravitation constant: $G=6.674*10^{-11}\frac{N*m^2}{kg^2}$

Estimate the gravitational force of the moon on a person of mass 101kg on the surface of the earth.

Possible Answers:

$2.45*10^{-3}N$

$8.54*10^{-3}N$

$6.18*10^{-3}N$

$3.36*10^{-3}N$

$8.81*10^{-3}N$

Correct answer:

$3.36*10^{-3}N$

Explanation:

Use the law of universal gravitation:

$F_{gravity}=G\frac{m_{1}m_{2}}{r^2}$

Convert to and plug in values:

$F_{gravity}=6.674*10^{-11}\frac{7.35*10^{22}*101}{(3.84*10^8)^2}$

$F_{gravity}=3.36*10^{-3}N$

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Example Question #681 : Ap Physics 1

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

Moon radius: 1.7*10^6m

$G=6.674*10^{-11}N*\frac{m^2}{kg^2}$

A spacecraft is orbiting 50km above the surface of the moon. Determine the velocity of the spacecraft.

Possible Answers:

$1700\frac{m}{s}$

$1200\frac{m}{s}$

$1300\frac{m}{s}$

$1900\frac{m}{s}$

$1500\frac{m}{s}$

Correct answer:

$1700\frac{m}{s}$

Explanation:

In an orbit, the centripetal force will be equal to the gravitational force:

$m_s\frac{v^2}{r}=G\frac{m_s*m_m}{r^2}$

Where:

is the distance to the moon's center. This will be the sum of the radius of the moon and the distance above the moon's center

is the linear velocity

m_s is the mass of the ship

m_m is the mass of the moon

is the universal gravity constant, $6.674*10^{-11}N*\frac{m^2}{kg^2}$

Simplify and plug in values:

${v^2}=6.67*10^{-11}\frac{7.35*10^{22}}{1.75*10^6}$

Solve for :

$v=1700\frac{m}{s}$

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Example Question #681 : Ap Physics 1

Mass of Earth: $5.97*10^{24}kg$

Mass of Moon: $7.25*10^{22}kg$

Distance between Earth and Moon: 3.84*10^8m

If a spaceship were to travel in a straight line towards the moon, at approximately what distance from the Earth would the forces of gravity due to the earth and moon be equal in magnitude?

Possible Answers:

2.41*10^8m

None of these

2.12*10^8m

1.11*10^8m

3.41*10^8m

Correct answer:

3.41*10^8m

Explanation:

Use the following equations to set up our calculation:

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

F_E=F_M

$R_{ship-earth}+R_{ship-moon}=R_{moon-earth}$

$G\frac{m_s m_E}{(R_{ship-earth})^2}=G\frac{m_s m_M}{(R_{ship-moon})^2}$

Combine equations and simplify:

$\frac{m_E}{(R_{ship-earth})^2}=\frac{m_M}{(R_{earth-moon}-R_{ship-earth})^2}$

Plug in values:

$\frac{5.97*10^{24}}{(R_{ship-earth})^2}=\frac{7.25*10^{22}}{(3.84*10^8-R_{ship-earth})^2}$

Solve for $R_{ship-earth}$ by converting to a quadratic equation:

$0=((\frac{7.25*10^{22}}{5.97*10^{24}})-1)R_{ship-earth}^2+2*3.84*10^8*R_{ship-earth}-(3.84*10^8)^2$

Use the quadratic formula to solve:

$R_{ship-earth}=3.41*10^8m$

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Example Question #41 : Universal Gravitation

Pluto radius: 1.186*10^6 m

Pluto mass: $1.3*10^{22} kg$

Determine the gravity constant, $g_{pluto}$ on the surface of Pluto.

Possible Answers:

$g_{pluto}=-2.4\frac{m}{s^2}$

$g_{pluto}=-.616\frac{m}{s^2}$

$g_{pluto}=-9.8\frac{m}{s^2}$

$g_{pluto}=-3.1\frac{m}{s^2}$

$g_{pluto}=-14.8\frac{m}{s^2}$

Correct answer:

$g_{pluto}=-.616\frac{m}{s^2}$

Explanation:

Set both forms of gravitational force equal to each other.

$m_{object}g_{pluto}=-G\frac{m_{pluto} m_{object}}{r^2}$

Simplify:

$g_{pluto}=-G\frac{m_{pluto}}{r^2}$

Plug in values:

$g_{pluto}=-6.674*10^{-11}\frac{1.3*10^{22}}{(1.186*10^6)^2}$

$g_{pluto}=-.616\frac{m}{s^2}$

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Example Question #41 : Universal Gravitation

How would the linear velocity of the Moon be different if it's mass was doubled? Assume that the distance to the Earth stayed the same.

Possible Answers:

Doubled

Unchanged

Halved

Quadrupled

Quartered

Correct answer:

Unchanged

Explanation:

Set centripetal force equal to gravitational force:

$m_M\frac{v^2}{r}=G\frac{m_M m_E}{r^2}$

$\frac{v^2}{r}=G\frac{m_E}{r^2}$

The mass of the Moon, m_M cancels out, thus, there is no effect.

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Example Question #43 : Universal Gravitation

If on earth you have a weight, , what would your new weight be if you were standing on a planet with the same mass as earth, but with half the radius?

Possible Answers:

$\frac{1}{2}$ your earth weight.

The same as your earth weight.

times your earth weight

$\frac{1}{2}$ your earth weight.

times your earth weight.

Correct answer:

times your earth weight.

Explanation:

Your weight is a function of how much force is pulling you down towards the planet, not just your mass. To find force when your standing on a different planet, you would need to use Newton's Law of Universal Gravitation.

$F = - \frac{G M}{r^{2}}$

To compare the forces from each planet, you would set this equation equal to itself.

$- \frac{G M_{earth}}{r_{earth}^{2}} = - \frac{G M_{planet}}{r_{planet}^{2}}$

We then cancel the common terms, in this formula that's the negative sign, (the gravitational constant), and the mass of the earth/planet (because they're the same). After that we can substitute the radius of the new planet for half of the earth radius.

$r_{planet}=\frac{1}{2}{}r_{earth}$

We need to remember that when we square the radius in the Law of Universal Gravitation, we also need to square the $\frac{1}{2}$ , making it $\frac{1}{4}$ . Because the $\frac{1}{4}$ is in the denominator, we take the inverse of it, so you would feel times more force standing on the new planet. Therefore you would have times as much weight.

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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 #641 : Newtonian Mechanics

Example Question #681 : Ap Physics 1

Example Question #641 : Newtonian Mechanics

Example Question #642 : Newtonian Mechanics

Example Question #643 : Newtonian Mechanics

Example Question #681 : Ap Physics 1

Example Question #681 : Ap Physics 1

Example Question #41 : Universal Gravitation

Example Question #41 : Universal Gravitation

Example Question #43 : Universal Gravitation

All AP Physics 1 Resources

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