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AP Physics 1 : Specific Forces

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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 #91 : Specific Forces

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:

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

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

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

None of these

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

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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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:

$790\frac{m}{s}$

$2506\frac{m}{s}$

$666\frac{m}{s}$

$1590\frac{m}{s}$

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 #62 : Universal Gravitation

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:

5.2N

10 N

32 N

.52 N

114 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 #63 : 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:

54240 N

542.4J

5424 N

5424 J

542.4N

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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AP Physics 1 : Specific Forces

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #51 : Universal Gravitation

Example Question #91 : Specific Forces

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

Example Question #61 : Universal Gravitation

Example Question #62 : Universal Gravitation

Example Question #63 : Universal Gravitation

All AP Physics 1 Resources

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