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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #75 : Forces

Two planets are 3*10^8m apart. If the first planet has a mass of $2.65*10^{20}kg$ and the second has a mass of $6.33*10^{26}kg$ , what is the acceleration on the larger planet?

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

Possible Answers:

$1.96*10^{-7}\frac{m}{s^2}$

$0.47\frac{m}{s^2}$

$5.1*10^6\frac{m}{s^2}$

$2.14\frac{m}{s^2}$

$5.1*10^3\frac{m}{s^2}$

Correct answer:

$1.96*10^{-7}\frac{m}{s^2}$

Explanation:

Remember that Newton's second law states that F=ma . The force acting upon the planet in question will be the force due to gravity. Once we find that, we can find the acceleration.

To solve for the force, use Newton's law of universal gravitation:

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

We are given the values for the mass of each planet, as well as the distance (radius) between them. Using these values and the gravitational constant, we can solve for the force of gravity.

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

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(2.65*10^{20}kg)*(6.33*10^{26}kg)}{(3*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.67745*10^{47}kg^2}{9*10^{16}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*1.864*10^{30}\frac{kg^2}{m^2}$

$F_G=1.24*10^{20}N$

Now that we know the force of gravity, we can use Newton's second law and the mass of the larger planet to solve for the acceleration of gravity.

F=ma

$1.24*10^{20}N=(6.33*10^{26}kg)*a$

$\frac{1.24*10^{20}N}{6.33*10^{26}kg}=a$

$1.96*10^{-7}\frac{m}{s^2}=a$

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Example Question #642 : Newtonian Mechanics

An astronaut has a mass of 71kg . He travels to a new planet and observes his weight is 510N on this planet's surface. If the radius of the planet is 6*10^7m , what is the mass of the planet?

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

Possible Answers:

$3.87*10^{26}kg$

$8.34*10^{31}kg$

$6.46*10^{18}kg$

$3.26*10^{22}kg$

$2.87*10^{-2}kg$

Correct answer:

$3.87*10^{26}kg$

Explanation:

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

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

Remember that the weight of the astronaut is the same as the gravitational force acting between the planet and the astronaut.

$F_g=mg=G\frac{m_1*m_2}{r^2}$

We are given the gravitational constant, the radius of the planet, the mass of the astronaut, and the magnitude of the force generated. Using these values in the universal gravitation equation, we can solve for the mass of the planet.

$510N=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(71kg)m_2}{(6*10^7m)^2}$

$\frac{510N}{6.67*10^{-11}\frac{m^3}{kg*s^2}}=\frac{(71kg)m_2}{(6*10^7m)^2}$

$7.65*10^{12}\frac{kg^2}{m^2}=\frac{(71kg)m_2}{3.6*10^{15}m^2}$

$(7.65*10^{12}\frac{kg^2}{m^2})(3.6*10^{15}m^2)=(71kg)m_2$

$2.75*10^{28}kg^2=(71kg)m_2$

$3.87*10^{26}=m_2$

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Example Question #31 : Universal Gravitation

Suppose that a person on Earth weighs 800N. If this person were to travel to a distant planet that had twice the density and the same radius of Earth, how much will the person weigh on this new planet?

$G=6.67384*10^{-11} \frac{m^{3}}{kg\cdot s^{2}}$

Possible Answers:

400N

200N

3200N

1600N

Correct answer:

1600N

Explanation:

We are given the weight of a person on Earth in units of Newtons, which means we can recognize this as a force. The force that is acting on this person is the force due to gravity, which can be represented by the following equation:

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

is the universal gravitational constant and is equal to $6.67384*10^{-11} \frac{m^{3}}{kg\cdot s^{2}}$

$m_{1}$ is the mass of object 1

$m_{2}$ is the mass of object 2

is the distance between the centers of the two objects

It's important to note that an object's mass will stay the same no matter where it is, but its weight will vary depending on where it is measured. Notice that when calculating the gravitational force, we need to consider the mass of two objects. If we set the mass of the Earth and the mass of the person in question as the two masses, we can rewrite the equation as:

$F_{Earth}=G\frac{m_{person}m_{Earth}}{r^{2}}=800N$

To calculate how much the person weighs on the new planet, we need to consider the information given - that the new planet is twice as dense as Earth. This means that for a given volume, the new planet will have twice as much mass as Earth. Furthermore, we know that the mass of the person stays the same since, as mentioned above, mass is constant no matter where it is measured. And, if we are considering a case where the volume is the same, then the distance between the centers should also be the same. Thus, we can calculate the new force as:

$F_{new\: planet}=G\frac{m_{person}(2m_{Earth})}{r^{2}}=2F_{Earth}=2\left ( 800 N\right )=1600\:N$

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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 #681 : Ap Physics 1

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:

$8.54*10^{-3}N$

$2.45*10^{-3}N$

$8.81*10^{-3}N$

$3.36*10^{-3}N$

$6.18*10^{-3}N$

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 #81 : Forces

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:

2.12*10^8m

None of these

1.11*10^8m

3.41*10^8m

2.41*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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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #75 : Forces

Example Question #642 : Newtonian Mechanics

Example Question #31 : Universal Gravitation

Example Question #641 : Newtonian Mechanics

Example Question #681 : Ap Physics 1

Example Question #641 : Newtonian Mechanics

Example Question #642 : Newtonian Mechanics

Example Question #681 : Ap Physics 1

Example Question #681 : Ap Physics 1

Example Question #81 : Forces

All AP Physics 1 Resources

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