Understanding Gravity and Gravitational Forces - AP Physics B

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Question

A certain planet has three times the radius of Earth and nine times the mass. How does the acceleration of gravity at the surface of this planet (ag) compare to the acceleration at the surface of Earth (g)?

Answer

The acceleration of gravity is given by the equation $a_{g} = \frac{GM}{r^{2}}$ , where G is constant.

For Earth, $a_{g} = \frac{GM_{earth}}{r_{earth}^{2}} = g$ .

For the new planet,

$a_{g} = \frac{G(9M_{earth})}{\left ( 3r_{earth} \right )^2} = \frac{G(9M_{earth})}{9r_{earth{}}^2} = \frac{GM_{earth}}{r_{earth}^{2}}} = g$ .

So, the acceleration is the same in both cases.

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Question

What is the acceleration due to gravity on a planet on which an object with a mass of 20.0kg has a weight of 270N?

Answer

Solve the following equation for acceleration, using the values given in the question.

$a = \frac{F}{m}$

$a=\frac{270N}{20kg}=13.5\frac{m}{s^2}$

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Question

The moon's distance from the center of the Earth was decreased by a multiple of three. How would this affect the gravitational force of the Earth on the moon?

Answer

The law of gravitation is written as $\small F = \frac{Gm_{1}m_{2}}{r^{2}}$ , with G being equal to $\small \small 6.6710^{-11}\frac{m^3}{kgs^2}$ .

Since the radius of the two masses acting on each other is squared, and is found in the denominator, a decrease in the radius by a multiple of three will cause a nine-fold increase in the gravitational force.

$\small F_1 = \frac{Gm_{1}m_{2}}{r^{2}}\rightarrow F_2=\frac{Gm_{1}m_{2}}{(\frac{1}{3}r)^{2}}=9F_1$

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Question

A ball is thrown horizontally off a cliff of height of $\small 50m$ with an initial velocity of $\small 15\frac{m}{s}$ . How far from the cliff will the ball land?

Answer

First we will find the time required for the ball to reach the ground. Since the ball is thrown horizontally, it has no initial vertical component. We use the following equation to solve for the total flight time:

$\Delta y=v_{iy}t+\frac{1}{2}at^{2}$

We are given the change in height, initial velocity, and acceleration. Using these values, we can solve for the time. Note that the change in height will be negative, since the ball is traveling downward.

$-50m= (0\frac{m}{s})t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-50m=(-4.9\frac{m}{s^2})t^2$

$t=\sqrt{\frac{-50m}{-4.9\frac{m}{s^2}}}$

Finally, we use the horizontal velocity to find the distance traveled in . Remember that the horizontal velocity remains constant during projectile motion.

$v=\frac{d}{t}$

$d=vt=(15\frac{m}{s})(3.19s)=47.9m$

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Question

A certain planet has three times the radius of Earth and nine times the mass. How does the acceleration of gravity at the surface of this planet (ag) compare to the acceleration at the surface of Earth (g)?

Answer

The acceleration of gravity is given by the equation $a_{g} = \frac{GM}{r^{2}}$ , where G is constant.

For Earth, $a_{g} = \frac{GM_{earth}}{r_{earth}^{2}} = g$ .

For the new planet,

$a_{g} = \frac{G(9M_{earth})}{\left ( 3r_{earth} \right )^2} = \frac{G(9M_{earth})}{9r_{earth{}}^2} = \frac{GM_{earth}}{r_{earth}^{2}}} = g$ .

So, the acceleration is the same in both cases.

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Question

What is the acceleration due to gravity on a planet on which an object with a mass of 20.0kg has a weight of 270N?

Answer

Solve the following equation for acceleration, using the values given in the question.

$a = \frac{F}{m}$

$a=\frac{270N}{20kg}=13.5\frac{m}{s^2}$

Compare your answer with the correct one above

Question

The moon's distance from the center of the Earth was decreased by a multiple of three. How would this affect the gravitational force of the Earth on the moon?

Answer

The law of gravitation is written as $\small F = \frac{Gm_{1}m_{2}}{r^{2}}$ , with G being equal to $\small \small 6.6710^{-11}\frac{m^3}{kgs^2}$ .

Since the radius of the two masses acting on each other is squared, and is found in the denominator, a decrease in the radius by a multiple of three will cause a nine-fold increase in the gravitational force.

$\small F_1 = \frac{Gm_{1}m_{2}}{r^{2}}\rightarrow F_2=\frac{Gm_{1}m_{2}}{(\frac{1}{3}r)^{2}}=9F_1$

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Question

A ball is thrown horizontally off a cliff of height of $\small 50m$ with an initial velocity of $\small 15\frac{m}{s}$ . How far from the cliff will the ball land?

Answer

First we will find the time required for the ball to reach the ground. Since the ball is thrown horizontally, it has no initial vertical component. We use the following equation to solve for the total flight time:

$\Delta y=v_{iy}t+\frac{1}{2}at^{2}$

We are given the change in height, initial velocity, and acceleration. Using these values, we can solve for the time. Note that the change in height will be negative, since the ball is traveling downward.

$-50m= (0\frac{m}{s})t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-50m=(-4.9\frac{m}{s^2})t^2$

$t=\sqrt{\frac{-50m}{-4.9\frac{m}{s^2}}}$

Finally, we use the horizontal velocity to find the distance traveled in . Remember that the horizontal velocity remains constant during projectile motion.

$v=\frac{d}{t}$

$d=vt=(15\frac{m}{s})(3.19s)=47.9m$

Compare your answer with the correct one above

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