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AP Physics 1 : Universal Gravitation

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Example Question #82 : Forces

An electronic scale is used to find the mass of a lead cube at sea level. The scale and lead cube are then transported to the top of a mountain, over $20000\textup{ ft}$ above sea level. How does the weight reading compare to the weight given at sea level?

Possible Answers:

It will be larger

It will be smaller

It is impossible to determine

None of these

It will be the same

Correct answer:

It will be smaller

Explanation:

The electronic scale measures based on the normal force the scale provides to the object. This in turn is based on the force of gravity on the object by the earth.

$F_G=-G\frac{m_1 m_2}{r^2}$

As height above sea level increases, as does , the distance to the center of the Earth. As increases, F_G decreases. This would decrease the normal force which would decrease the reading on the scale.

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Example Question #82 : Forces

$G=6.67*10^{-11}\frac{N*m^2}{kg^2}$

Mass of moon: $7.3*10^{22}kg$

Radius of moon: 1737km

A spring of rest length 15cm is placed upright on the moon. A mass of 15kg is gently placed on top and the spring contracts by 1cm . Determine the spring constant.

Possible Answers:

$\frac{3388N}{m}$

$\frac{3751N}{m}$

$\frac{4449N}{m}$

None of these

$\frac{2415N}{m}$

Correct answer:

$\frac{2415N}{m}$

Explanation:

First, estimate the acceleration due to gravity close to the moon's surface:

F=ma

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

Combining equations and solving for the acceleration:

$a_{mass}=-G\frac{m_{moon}}{r_{moon}^2}$

Converting to and pugging in values:

$a_{mass}=-6.67*10^{-11}\frac{7.3*10^{22}}{(1737*10^3)^2}$

$a_{mass}=-1.61\frac{m}{s^2}$

Using

$F_{total}=F_1+F_2+...$

$F_{gravity}=ma_{gravity}$

$F_{total}=kx+ma_{gravity}$

Solving for

$k=\frac{F_{total}-ma_{gravity}}{x}$

Converting to and plugging in values:

$k=\frac{0-15*-1.61}{.01}$

(Since the mass is at rest, the acceleration and thus the net force is zero)

$k=\frac{2415N}{m}$

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Example Question #652 : Newtonian Mechanics

A rope is used to accelerate a 5kg mass upwards at $2\frac{m}{s^2}$ . Determine the tension in the rope.

Possible Answers:

59N

90N

202N

18N

None of these

Correct answer:

59N

Explanation:

Using

F=ma

$F_{total}=F_1+F_2+...$

Combining equations:

$ma=F_{tension}+mg$

Solving for $F_{tension}$ and plugging in values:

$F_{tension}=5*2-5*(-9.8)$

$F_{tension}=59N$

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Example Question #51 : Universal Gravitation

Mass of Earth: $5.972*10^{24} kg$

Universal gravitation constant: $6.67*10^{-11}\frac{N\cdot m^2}{kg^2}$

Radius of earth: 6371km

Determine the magnitude of gravitational force by the earth on a 110kg astronaut 200km above the surface of the earth.

Possible Answers:

1015N

1455N

987N

1110N

1561N

Correct answer:

1015N

Explanation:

Finding total distance from center of Earth to astronaut:

200km+6371km=6571km

Converting to meters

6.571*10^6m

Using Universal Gravitation equation:

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

Plugging in values:

$F_G=-6.67*10^{-11}\frac{5.972*10^{24}*110}{(6.571*10^6)^2}$

|F_G|=1015N

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Example Question #52 : Universal Gravitation

Mass of Jupiter: $1.89*10^{27}kg$

Universal gravitation constant: $6.67*10^{-11}\frac{N*m^2}{kg^2}$

Radius of Jupiter: 69911km

A marble is placed 100000km from the surface of Jupiter. Determine the acceleration due to the gravity of Jupiter.

Possible Answers:

$4.52*10^{-2}\frac{m}{s^2}$

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

None of these

$4.52*10^{2}\frac{m}{s^2}$

$3.55*10^{-2}\frac{m}{s^2}$

Correct answer:

$4.52*10^{-2}\frac{m}{s^2}$

Explanation:

Using

F=ma

and

$|F_G|=G\frac{m_1 m_2}{r^2}$

Combining equations

$m_1a=G\frac{m_1m_2}{r^2}$

Solving for

$a=G\frac{m_2}{r^2}$

Plugging in values:

$a=6.67*10^{-11}\frac{1.89*10^{27}}{(1670*10^6)^2}$

$a=4.52*10^{-2}\frac{m}{s^2}$

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Example Question #53 : Universal Gravitation

$G=6.67*10^{-11}\frac{N*m^2}{kg^2}$

Mass of Pluto: $1.31*10^{22}kg$

Radius of Pluto: $1.19*10^{6}m$

A spring of rest length 27cm is placed upright on Pluto. A mass of 27kg is gently placed on top and the spring contracts by 4cm . Determine the spring constant.

Possible Answers:

$\frac{416N}{m}$

$\frac{611N}{m}$

$\frac{844N}{m}$

$\frac{557N}{m}$

None of these

Correct answer:

$\frac{416N}{m}$

Explanation:

First, estimate the acceleration due to gravity close to Pluto's surface:

F=ma

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

Combining equations and solving for the acceleration:

$a_{mass}=-G\frac{m_{pluto}}{r_{pluto}^2}$

Plugging in values:

$a_{mass}=-6.67*10^{-11}\frac{1.31*10^{22}}{(1.19*10^6)^2}$

$a_{mass}=-.617\frac{m}{s^2}$

Using

$F_{total}=F_1+F_2+...$

$F_{gravity}=ma_{gravity}$

$F_{total}=kx+ma_{gravity}$

Solving for

$k=\frac{F_{total}-ma_{gravity}}{x}$

Converting to and plugging in values:

$k=\frac{0-27*(-.617)}{.04}$

(Since the mass is at rest, the acceleration and thus the net force is zero)

$k=\frac{416N}{m}$

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Example Question #661 : Newtonian Mechanics

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

Mass of Mars: $639*10^{21}kg$

Radius of Mars: $3.39*10^{6}m$

A spring of rest length 18cm is placed upright on Mars. A mass of 65kg is gently placed on top and the spring contracts by 5.5cm . Determine the spring constant.

Possible Answers:

$k=\frac{4014N}{m}$

$k=\frac{3985N}{m}$

None of these

$k=\frac{3751N}{m}$

$k=\frac{4385N}{m}$

Correct answer:

$k=\frac{4385N}{m}$

Explanation:

First, estimate the acceleration due to gravity close to the martian surface:

F=ma

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

Combining equations and solving for the acceleration:

$a_{mass}=-G\frac{m_{mars}}{r_{mars}^2}$

Plugging in values:

$a_{mass}=-6.67*10^{-11}\frac{639*10^{21}}{(3.39*10^6)^2}$

$a_{mass}=-3.71\frac{m}{s^2}$

Using

$F_{total}=F_1+F_2+...$

$F_{gravity}=ma_{gravity}$

$F_{total}=kx+ma_{gravity}$

Solving for

$k=\frac{F_{total}-ma_{gravity}}{x}$

Converting to and plugging in values:

$k=\frac{0-65*(-3.71)}{.055}$

(Since the mass is at rest, the acceleration and thus the net force is zero)

$k=\frac{4385N}{m}$

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Example Question #662 : Newtonian Mechanics

Determine the surface gravitational constant for a roughly spherical asteroid of mass $6.3*10^{20}kg$ and a radius of 250km .

Possible Answers:

$.782\frac{m}{s}$

$.672\frac{m}{s}$

None of these

$.611\frac{m}{s}$

$.638\frac{m}{s}$

Correct answer:

$.672\frac{m}{s}$

Explanation:

Setting universal gravitation equal to the surface gravitational approximation

$-G\frac{m_1m_2}{r^2}=-m_2*g_a$

Where g_a will be the surface gravitational acceleration constant

Solving for g_a

$g_a=G\frac{m_1}{r^2}$

Plugging in values

$g_a=6.67*10^{-11}\frac{6.3*10^{20}}{250,000^2}$

$g_a=.672\frac{m}{s}$

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Example Question #701 : Ap Physics 1

Determine the surface gravitational constant for a roughly spherical asteroid of mass $1.1*10^{20}kg$ and a radius of 200km .

Possible Answers:

None of these

$.183\frac{m}{s^2}$

$9.8\frac{m}{s^2}$

$3.33\frac{m}{s^2}$

$5.41\frac{m}{s^2}$

Correct answer:

$.183\frac{m}{s^2}$

Explanation:

Setting universal gravitation equal to the gravitational approximation

$-G\frac{m_1m_2}{r^2}=-m_2*g_a$

Where g_a will be the surface gravitational acceleration constant

Solving for g_a

$g_a=G\frac{m_1}{r^2}$

Converting to and plugging in values

$g_a=6.67*10^{-11}\frac{1.1*10^{20}}{200,000^2}$

$g_a=.183\frac{m}{s^s}$

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Example Question #664 : Newtonian Mechanics

Astronauts have recently detected a new exoplanet, Zina. The mass of the planet is twice that of Earth, and the radius is three times larger than Earth's radius. What fraction of our gravity is experienced on this planet?

Possible Answers:

$\frac{2}{3}$

$\frac{2}{9}$

$\frac{3}{2}$

$\frac{9}{2}$

Same gravity because gravity is constant.

Correct answer:

$\frac{2}{9}$

Explanation:

The gravity equation is:

$g=\frac{GM}{R^2}$ , where is the universal gravitation constant, is the mass of the planet, and is the radius of planet. To find a ratio between the gravity on the planet Zina and our planet, we need to plug in the relevant information:

$\frac{g_{Zina}}{g_{Earth}}=\frac{\frac{GM_{Zina}}{R^2_{Zina}}}{\frac{GM_{Earth}}{R^2_{Earth}}}=\frac{\frac{G*2*M_{Earth}}{3^2*R^2_{Earth}}}{\frac{GM_{Earth}}{R^2_{Earth}}}$

$=\frac{\frac{G*2}{3^2}}{\frac{G*1}{1}}=\frac{2}{9}.$

Remember, we are doing a ratio, so if we reference everything in terms of Earth numbers, the bottom fraction becomes much simpler and cancels out with the top fraction. Also, the gravitational constant is constant, so it cancels out. Therefore, the only important information are the ratios the mass and radius differ from Earth'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 #82 : Forces

Example Question #82 : Forces

Example Question #652 : Newtonian Mechanics

Example Question #51 : Universal Gravitation

Example Question #52 : Universal Gravitation

Example Question #53 : Universal Gravitation

Example Question #661 : Newtonian Mechanics

Example Question #662 : Newtonian Mechanics

Example Question #701 : Ap Physics 1

Example Question #664 : Newtonian Mechanics

All AP Physics 1 Resources

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