Gravity and Weight - MCAT Physical

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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

A hypothetical planet has a radius equal to twice that of Earth, with the same mass as Earth. How much would a person weigh on the surface of this hypothetical planet if they weighed 1000N on earth?

Answer

The force due to gravity on any object can be given by the equation below.

$F = \frac{GMm}{r^{2}}$

is the gravitational constant, is the mass of the earth, is the mass of the object, and is the distance between the center of each object.

In our question, the only value to change is the radius of the new planet; both masses and remain constant. The effect of doubling the radius on the force is given below.

$F_2=\frac{GMm}{(2)^2}=\frac{1}{4}F_1$

The person's weight on the new planet would be one-fourth their weight on Earth.

$\frac{1}{4}1000N=250N$

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Question

A person stands on a scale in an elevator. When the elevator is moving upwards at a constant velocity of 0.5m/s, the scale reads 500N. If the elevator then slows down, with a deceleration of 0.5m/s2, what is the new reading?

Answer

It may be helpful to start with a free-body diagram showing the forces acting on the person. We have the gravitational force, Fg, downwards, and the normal force of the scale, Fn, upwards.

Use these to write the net force equation.

$\small F_{net} = ma = F_{n}-F_{g}$

First we need to solve for the person’s mass, m. When the elevator is moving at a constant rate, we are given that Fn = 500N, and we can solve for m.

$\small 0=F_{n}-mg = 500-mg$ .

$\small m=\frac{500N}{10\frac{m}{s^{2}}}=50kg$

When the elevator comes to rest, then we have an acceleration of -0.5 m/s2 (downwards acceleration). Plugging this in, we can solve for the new Fn.

$\small (50kg)(-0.5 \frac{m}{s^{2}})=F_{n}-(50kg)(10\frac{m}{s^{2}})$

$\small F_{n}=500-25 = 475 N$

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Question

What is the tension in a rope that is used to pull a box straight upwards with an acceleration of $1\frac{m}{s^2}$ ?

Answer

For the problem we need to understand Newton's second law: . The net upward forces and net downward forces must equal the product of mass and acceleration. Since the box is traveling upwards with an acceleration of 1m/s2 we can indicate upward forces as positive and downward forces as negative. Indicating T as tension and Fg as the weight of the box, we can find the value of tension with the equation .

$T=ma+F_g=(20kg)(1\frac{m}{s^2})+(20kg)(10\frac{m}{s^2})=220N$

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Question

A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

What is the apparent weight of the weights on the platform as they fall after the rope has been cut?

Answer

First, we need to determine what the apparent weight actually is. If we draw a free body diagram of the weights, we note that the force due to gravity acts downward and the force of the platform on the weights (the normal force) acts upward.

The normal force, FN in the above diagram, is what a scale reads, and is thus the apparent weight. Finding the normal force is not as easy as equating it to the weight because the system is in free-fall, and thus accelerating. We can use Newton’s second law and equate it to acceleration to find the normal force.

Assuming downward is the positive y direction, we can solve for FN.

mg – FN = ma

FN = mg – ma = 0 N

Logically, this makes sense. If an object is in free fall, the acceleration of the system is the same as the acceleration due to gravity.

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Question

Planet X has mass $\small M$ and radius $\small R$ , and the gravitational acceleration on its surface is $\small 15\frac{m}{s^2}$ . What is the gravitational acceleration at the location of a satellite orbiting at a distance $\small 3R$ from the surface of planet X?

Answer

Gravitational acceleration is related to distance via the equation:

$a=\frac{GM}{r^2}$

$\small G$ is the gravitational constant, $\small M$ is the mass of the attracting object, and $\small r$ is the distance from its center.

In this case, the initial distance (on the surface) is $\small 1R$ from the center of planet X, and the distance of the satellite is $\small 4R$ from the center. We know that gravitational acceleration is proportional to the distance squared, and we know the acceleration at the surface. Using these values, we can solve for the acceleration on the satellite.

$a_{surface}=\frac{GM}{R^2}=15\frac{m}{s^2}$

$a_{satellite}=\frac{GM}{(4R)^2}=\frac{GM}{16R^2}$

$a_{satellite}=\frac{1}{16}\frac{GM}{R^2}$

By expanding the equation for the acceleration on the satellite, we can see that it is equal to one-sixteenth the acceleration at the surface, based on our original equation. Substitute the value of the surface acceleration to get the final answer.

$a_{satellite}=\frac{1}{16}(a_{surface})=\frac{1}{16}(15\frac{m}{s^2})$

$a_{satellite}=\frac{15}{16}\frac{m}{s^2}$

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Question

An astronaut standing on a scale on the moon observes that he weighs $\small 100N$ . If the acceleration due to gravity on the moon's is one-sixth of its value on Earth's surface, what is the astronaut's mass on the earth?

Answer

To relate force and mass, we use Newton's second law:

We are given his weight (force) on the moon, and we are told the relative gravitational acceleration on the moon. Using these values, we can find the astronaut's mass.

$F_m=100N\ \text{and}\ a_m=\frac{1}{6}g$

$100N=m(\frac{1}{6}g)$

$100N=m(\frac{1}{6}*9.8\frac{m}{s^2})$

$m=\frac{100N}{1.63\frac{m}{s^2}}=61.22kg$

Since mass is the same regardless of gravitational acceleration, this is the same as the astronaut's mass on the earth.

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Question

A $\small 100g$ block lies upon a frictionless surface. A string is attached to the right side of the block, passed over a pulley, and then attached to a $\small 200g$ mass suspended by the string. When the block is held still, what is the tension in the string?

Answer

The block cannot provide any frictional force to the left because it is resting on a frictionless table. Whether the block is allowed to move or not, the tension force in the string is all generated by the suspended $\small 200g$ mass. The force in the string is equal and opposite to the gravitational force acting on the suspended mass.

$F_T=-(0.2kg)(-9.8\frac{m}{s^2})=1.96N$

The system is in equilibrium when no parts are in motion. Alternatively stated, the upward tension in the string is equal and opposite to the downwards force on the mass. If the block is not allowed to move as more mass is added, at some point the tensile force in the string will exceed its mechanical limits, and the string will break.

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Question

A ball is thrown vertically with an initial velocity, , and returns to its original position after time . How would the value of be affected if the ball were thrown in the same manner on the moon, where gravitational acceleration is one-sixth the gravitational acceleration on Earth?

Answer

To solve this question, we will need to use the equation for acceleration:

$a=\frac{v_2-v_1}{t}$

In this case, the initial velocity will be equal to the final velocity, but opposite in magnitude. The initial velocity is in the upward direction, while the final velocity is downward.

Plug this value into the equation for acceleration.

$a=\frac{-2v}{t}$

The velocity value is constant, regardless of the planet. Substitute the acceleration for each planet to determine the change in the time variable.

$a_E=g=\frac{-2v}{t}\rightarrow t_{E}=\frac{-2v}{g}$

$a_M=\frac{g}{6}=\frac{-2v}{t}\rightarrow t_M=6\frac{-2v}{g}$

We can see that the time of flight on the moon is equal to six times the time of flight on Earth.

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Question

A sphere of mass is separated from another sphere of mass by a distance . What is the gravitational force between the two spheres?

Answer

The force between two objects is calculated by:

$F_G=\frac{Gm_{1}m_{2}}{d^2}$

Use our given variables:

$F_G=\frac{G(4M)(M)}{D^2}=4\frac{GM^2}{D^2}$

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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.

Compare your answer with the correct one above

Question

A hypothetical planet has a radius equal to twice that of Earth, with the same mass as Earth. How much would a person weigh on the surface of this hypothetical planet if they weighed 1000N on earth?

Answer

The force due to gravity on any object can be given by the equation below.

$F = \frac{GMm}{r^{2}}$

is the gravitational constant, is the mass of the earth, is the mass of the object, and is the distance between the center of each object.

In our question, the only value to change is the radius of the new planet; both masses and remain constant. The effect of doubling the radius on the force is given below.

$F_2=\frac{GMm}{(2)^2}=\frac{1}{4}F_1$

The person's weight on the new planet would be one-fourth their weight on Earth.

$\frac{1}{4}1000N=250N$

Compare your answer with the correct one above

Question

A person stands on a scale in an elevator. When the elevator is moving upwards at a constant velocity of 0.5m/s, the scale reads 500N. If the elevator then slows down, with a deceleration of 0.5m/s2, what is the new reading?

Answer

It may be helpful to start with a free-body diagram showing the forces acting on the person. We have the gravitational force, Fg, downwards, and the normal force of the scale, Fn, upwards.

Use these to write the net force equation.

$\small F_{net} = ma = F_{n}-F_{g}$

First we need to solve for the person’s mass, m. When the elevator is moving at a constant rate, we are given that Fn = 500N, and we can solve for m.

$\small 0=F_{n}-mg = 500-mg$ .

$\small m=\frac{500N}{10\frac{m}{s^{2}}}=50kg$

When the elevator comes to rest, then we have an acceleration of -0.5 m/s2 (downwards acceleration). Plugging this in, we can solve for the new Fn.

$\small (50kg)(-0.5 \frac{m}{s^{2}})=F_{n}-(50kg)(10\frac{m}{s^{2}})$

$\small F_{n}=500-25 = 475 N$

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Question

What is the tension in a rope that is used to pull a box straight upwards with an acceleration of $1\frac{m}{s^2}$ ?

Answer

For the problem we need to understand Newton's second law: . The net upward forces and net downward forces must equal the product of mass and acceleration. Since the box is traveling upwards with an acceleration of 1m/s2 we can indicate upward forces as positive and downward forces as negative. Indicating T as tension and Fg as the weight of the box, we can find the value of tension with the equation .

$T=ma+F_g=(20kg)(1\frac{m}{s^2})+(20kg)(10\frac{m}{s^2})=220N$

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Question

A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

What is the apparent weight of the weights on the platform as they fall after the rope has been cut?

Answer

First, we need to determine what the apparent weight actually is. If we draw a free body diagram of the weights, we note that the force due to gravity acts downward and the force of the platform on the weights (the normal force) acts upward.

The normal force, FN in the above diagram, is what a scale reads, and is thus the apparent weight. Finding the normal force is not as easy as equating it to the weight because the system is in free-fall, and thus accelerating. We can use Newton’s second law and equate it to acceleration to find the normal force.

Assuming downward is the positive y direction, we can solve for FN.

mg – FN = ma

FN = mg – ma = 0 N

Logically, this makes sense. If an object is in free fall, the acceleration of the system is the same as the acceleration due to gravity.

Compare your answer with the correct one above

Question

Planet X has mass $\small M$ and radius $\small R$ , and the gravitational acceleration on its surface is $\small 15\frac{m}{s^2}$ . What is the gravitational acceleration at the location of a satellite orbiting at a distance $\small 3R$ from the surface of planet X?

Answer

Gravitational acceleration is related to distance via the equation:

$a=\frac{GM}{r^2}$

$\small G$ is the gravitational constant, $\small M$ is the mass of the attracting object, and $\small r$ is the distance from its center.

In this case, the initial distance (on the surface) is $\small 1R$ from the center of planet X, and the distance of the satellite is $\small 4R$ from the center. We know that gravitational acceleration is proportional to the distance squared, and we know the acceleration at the surface. Using these values, we can solve for the acceleration on the satellite.

$a_{surface}=\frac{GM}{R^2}=15\frac{m}{s^2}$

$a_{satellite}=\frac{GM}{(4R)^2}=\frac{GM}{16R^2}$

$a_{satellite}=\frac{1}{16}\frac{GM}{R^2}$

By expanding the equation for the acceleration on the satellite, we can see that it is equal to one-sixteenth the acceleration at the surface, based on our original equation. Substitute the value of the surface acceleration to get the final answer.

$a_{satellite}=\frac{1}{16}(a_{surface})=\frac{1}{16}(15\frac{m}{s^2})$

$a_{satellite}=\frac{15}{16}\frac{m}{s^2}$

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Question

An astronaut standing on a scale on the moon observes that he weighs $\small 100N$ . If the acceleration due to gravity on the moon's is one-sixth of its value on Earth's surface, what is the astronaut's mass on the earth?

Answer

To relate force and mass, we use Newton's second law:

We are given his weight (force) on the moon, and we are told the relative gravitational acceleration on the moon. Using these values, we can find the astronaut's mass.

$F_m=100N\ \text{and}\ a_m=\frac{1}{6}g$

$100N=m(\frac{1}{6}g)$

$100N=m(\frac{1}{6}*9.8\frac{m}{s^2})$

$m=\frac{100N}{1.63\frac{m}{s^2}}=61.22kg$

Since mass is the same regardless of gravitational acceleration, this is the same as the astronaut's mass on the earth.

Compare your answer with the correct one above

Question

A $\small 100g$ block lies upon a frictionless surface. A string is attached to the right side of the block, passed over a pulley, and then attached to a $\small 200g$ mass suspended by the string. When the block is held still, what is the tension in the string?

Answer

The block cannot provide any frictional force to the left because it is resting on a frictionless table. Whether the block is allowed to move or not, the tension force in the string is all generated by the suspended $\small 200g$ mass. The force in the string is equal and opposite to the gravitational force acting on the suspended mass.

$F_T=-(0.2kg)(-9.8\frac{m}{s^2})=1.96N$

The system is in equilibrium when no parts are in motion. Alternatively stated, the upward tension in the string is equal and opposite to the downwards force on the mass. If the block is not allowed to move as more mass is added, at some point the tensile force in the string will exceed its mechanical limits, and the string will break.

Compare your answer with the correct one above

Question

A ball is thrown vertically with an initial velocity, , and returns to its original position after time . How would the value of be affected if the ball were thrown in the same manner on the moon, where gravitational acceleration is one-sixth the gravitational acceleration on Earth?

Answer

To solve this question, we will need to use the equation for acceleration:

$a=\frac{v_2-v_1}{t}$

In this case, the initial velocity will be equal to the final velocity, but opposite in magnitude. The initial velocity is in the upward direction, while the final velocity is downward.

Plug this value into the equation for acceleration.

$a=\frac{-2v}{t}$

The velocity value is constant, regardless of the planet. Substitute the acceleration for each planet to determine the change in the time variable.

$a_E=g=\frac{-2v}{t}\rightarrow t_{E}=\frac{-2v}{g}$

$a_M=\frac{g}{6}=\frac{-2v}{t}\rightarrow t_M=6\frac{-2v}{g}$

We can see that the time of flight on the moon is equal to six times the time of flight on Earth.

Compare your answer with the correct one above

Question

A sphere of mass is separated from another sphere of mass by a distance . What is the gravitational force between the two spheres?

Answer

The force between two objects is calculated by:

$F_G=\frac{Gm_{1}m_{2}}{d^2}$

Use our given variables:

$F_G=\frac{G(4M)(M)}{D^2}=4\frac{GM^2}{D^2}$

Compare your answer with the correct one above

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