Universal Gravitation - MCAT Physical

Card 0 of 49

Question

Two objects with masses of M and m, sit r distance apart. What will be the effect on the gravitational force between them if the masses are changed to 2M and 3m?

It will increase 36-fold.

It will increase 6-fold.

It will increase 3-fold.

It will not change unless r is changed.

None of the above.

Answer

Choice 2 is correct. The formula for gravitational force, G, is $\dpi{100} \small G=\frac{k\left ( m_{1} \right )\left ( m_{2} \right )}{r^{2}}$ , where k is a constant. The effect of doubling one mass and trebling the other is multiplicative, $\dpi{100} \small 2\times 3$ , so the answer is six-fold. The question attempts to confuse the respondent by forcing them to recall that the element of the equation which is squared is distance between objects.

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Question

An object has a mass of 50kg and a weight of 500N when it is resting on the surface of the Earth. If it is moved to a height equal to three times the Earth’s radius, what is the object’s new weight?

Answer

You should be familiar with the following equation for the force of gravity.

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

To solve this problem, recognize that weight (Fg) is proportional to the inverse square of the radius. When the object was a distance of r (Earth’s radius) it had a weight of 500N. Now, the object is at a distance of 4r (radius of Earth plus the 3r distance that the object is moved to). With the proportion described above, we can see that the force is decreased by a factor of (4)2.

$\frac{1}{4^{2}}(500 N) = \frac{1}{16}(500 N) = 30 N$

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.

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

Compare your answer with the correct one above

Question

Using Newton's Law of Universal Gravitation equation, which of the following expressions is equal to the local gravitational acceleration on Earth?

Answer

On earth, .

The law of universal gravitation is equal to $F = \frac{GM_{earth}m}{R_{earth}^{2}}$ .

We can set these equations equal to one another and isolate by dividing both sides by , the mass of an object on Earth.

$mg=\frac{GM_{earth}m}{R^2_{earth}}$

$g=\frac{GM_{earth}}{R^2_{earth}}$

Using the mass of the Earth, the radius of the Earth, and the gravitational constant, , we get a value of approximately $9.8\frac{m}{s^2}$ if we solve for .

Compare your answer with the correct one above

Question

A body with mass $\small m$ is situated $\small 3.0$ meters from a second body with a mass of $\small 2m$ . What will be the effect on gravitational attraction of moving one body so that it is only $\small 1.74$ meters from the other body?

Answer

Gravity is essentially a property of mass, and the force of gravitational attraction between two bodies is given by the formula:

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

In our scenario, the masses remain the same, and of course $\small G$ is a constant, so the only thing that changes is the denominator.

$F_{g1} = G\frac{m_{1}m_{2}}{(3.0)^{2}}=G\frac{m_{1}m_{2}}{9}$

$F_{g2} = G\frac{m_{1}m_{2}}{(1.74)^{2}}=G\frac{m_{1}m_{2}}{3.03}$

$\frac{9}{3.03} \cong 3$

$F_{g2}\approx3F_{g1}$

Compare your answer with the correct one above

Question

Which of the following changes would increase a satellite's orbital speed?

Answer

We know , and the force of gravity is:

$F_{grav}=\frac{GMm}{R^{2}}$

Also, the equation for uniform circular motion, such as a satellite in orbit is:

$a=\frac{v^{2}}{R}$

Set $F_{grav}=ma$ and substitute the acceleration due to circular motion into the equation. Solve for velocity.

$v = \sqrt{\frac{GM}{R}}$

This indicates that the only variables that affect the orbital speed are orbital radius and the mass of the Earth.

Compare your answer with the correct one above

Question

Which of the following must be true for a block to be considered weightless?

Answer

One way an object can be considered "weightless," is if it is accelerating downward at the same rate that it would be if it were free falling. This is why objects in an elevator whose cable had been cut would be "weightless." Therefore the answer is just for the block to be accelerating in the same magnitude and direction as the acceleration due to gravity, which is $9.8\frac{m}{s^2}$

Compare your answer with the correct one above

Question

Two objects with masses of M and m, sit r distance apart. What will be the effect on the gravitational force between them if the masses are changed to 2M and 3m?

It will increase 36-fold.

It will increase 6-fold.

It will increase 3-fold.

It will not change unless r is changed.

None of the above.

Answer

Choice 2 is correct. The formula for gravitational force, G, is $\dpi{100} \small G=\frac{k\left ( m_{1} \right )\left ( m_{2} \right )}{r^{2}}$ , where k is a constant. The effect of doubling one mass and trebling the other is multiplicative, $\dpi{100} \small 2\times 3$ , so the answer is six-fold. The question attempts to confuse the respondent by forcing them to recall that the element of the equation which is squared is distance between objects.

Compare your answer with the correct one above

Question

An object has a mass of 50kg and a weight of 500N when it is resting on the surface of the Earth. If it is moved to a height equal to three times the Earth’s radius, what is the object’s new weight?

Answer

You should be familiar with the following equation for the force of gravity.

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

To solve this problem, recognize that weight (Fg) is proportional to the inverse square of the radius. When the object was a distance of r (Earth’s radius) it had a weight of 500N. Now, the object is at a distance of 4r (radius of Earth plus the 3r distance that the object is moved to). With the proportion described above, we can see that the force is decreased by a factor of (4)2.

$\frac{1}{4^{2}}(500 N) = \frac{1}{16}(500 N) = 30 N$

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.

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

Compare your answer with the correct one above

Question

Using Newton's Law of Universal Gravitation equation, which of the following expressions is equal to the local gravitational acceleration on Earth?

Answer

On earth, .

The law of universal gravitation is equal to $F = \frac{GM_{earth}m}{R_{earth}^{2}}$ .

We can set these equations equal to one another and isolate by dividing both sides by , the mass of an object on Earth.

$mg=\frac{GM_{earth}m}{R^2_{earth}}$

$g=\frac{GM_{earth}}{R^2_{earth}}$

Using the mass of the Earth, the radius of the Earth, and the gravitational constant, , we get a value of approximately $9.8\frac{m}{s^2}$ if we solve for .

Compare your answer with the correct one above

Question

A body with mass $\small m$ is situated $\small 3.0$ meters from a second body with a mass of $\small 2m$ . What will be the effect on gravitational attraction of moving one body so that it is only $\small 1.74$ meters from the other body?

Answer

Gravity is essentially a property of mass, and the force of gravitational attraction between two bodies is given by the formula:

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

In our scenario, the masses remain the same, and of course $\small G$ is a constant, so the only thing that changes is the denominator.

$F_{g1} = G\frac{m_{1}m_{2}}{(3.0)^{2}}=G\frac{m_{1}m_{2}}{9}$

$F_{g2} = G\frac{m_{1}m_{2}}{(1.74)^{2}}=G\frac{m_{1}m_{2}}{3.03}$

$\frac{9}{3.03} \cong 3$

$F_{g2}\approx3F_{g1}$

Compare your answer with the correct one above

Question

Which of the following changes would increase a satellite's orbital speed?

Answer

We know , and the force of gravity is:

$F_{grav}=\frac{GMm}{R^{2}}$

Also, the equation for uniform circular motion, such as a satellite in orbit is:

$a=\frac{v^{2}}{R}$

Set $F_{grav}=ma$ and substitute the acceleration due to circular motion into the equation. Solve for velocity.

$v = \sqrt{\frac{GM}{R}}$

This indicates that the only variables that affect the orbital speed are orbital radius and the mass of the Earth.

Compare your answer with the correct one above

Question

Which of the following must be true for a block to be considered weightless?

Answer

One way an object can be considered "weightless," is if it is accelerating downward at the same rate that it would be if it were free falling. This is why objects in an elevator whose cable had been cut would be "weightless." Therefore the answer is just for the block to be accelerating in the same magnitude and direction as the acceleration due to gravity, which is $9.8\frac{m}{s^2}$

Compare your answer with the correct one above

Question

Two objects with masses of M and m, sit r distance apart. What will be the effect on the gravitational force between them if the masses are changed to 2M and 3m?

It will increase 36-fold.

It will increase 6-fold.

It will increase 3-fold.

It will not change unless r is changed.

None of the above.

Answer

Choice 2 is correct. The formula for gravitational force, G, is $\dpi{100} \small G=\frac{k\left ( m_{1} \right )\left ( m_{2} \right )}{r^{2}}$ , where k is a constant. The effect of doubling one mass and trebling the other is multiplicative, $\dpi{100} \small 2\times 3$ , so the answer is six-fold. The question attempts to confuse the respondent by forcing them to recall that the element of the equation which is squared is distance between objects.

Compare your answer with the correct one above

Question

An object has a mass of 50kg and a weight of 500N when it is resting on the surface of the Earth. If it is moved to a height equal to three times the Earth’s radius, what is the object’s new weight?

Answer

You should be familiar with the following equation for the force of gravity.

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

To solve this problem, recognize that weight (Fg) is proportional to the inverse square of the radius. When the object was a distance of r (Earth’s radius) it had a weight of 500N. Now, the object is at a distance of 4r (radius of Earth plus the 3r distance that the object is moved to). With the proportion described above, we can see that the force is decreased by a factor of (4)2.

$\frac{1}{4^{2}}(500 N) = \frac{1}{16}(500 N) = 30 N$

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.

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

Compare your answer with the correct one above

Question

Using Newton's Law of Universal Gravitation equation, which of the following expressions is equal to the local gravitational acceleration on Earth?

Answer

On earth, .

The law of universal gravitation is equal to $F = \frac{GM_{earth}m}{R_{earth}^{2}}$ .

We can set these equations equal to one another and isolate by dividing both sides by , the mass of an object on Earth.

$mg=\frac{GM_{earth}m}{R^2_{earth}}$

$g=\frac{GM_{earth}}{R^2_{earth}}$

Using the mass of the Earth, the radius of the Earth, and the gravitational constant, , we get a value of approximately $9.8\frac{m}{s^2}$ if we solve for .

Compare your answer with the correct one above

Question

A body with mass $\small m$ is situated $\small 3.0$ meters from a second body with a mass of $\small 2m$ . What will be the effect on gravitational attraction of moving one body so that it is only $\small 1.74$ meters from the other body?

Answer

Gravity is essentially a property of mass, and the force of gravitational attraction between two bodies is given by the formula:

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

In our scenario, the masses remain the same, and of course $\small G$ is a constant, so the only thing that changes is the denominator.

$F_{g1} = G\frac{m_{1}m_{2}}{(3.0)^{2}}=G\frac{m_{1}m_{2}}{9}$

$F_{g2} = G\frac{m_{1}m_{2}}{(1.74)^{2}}=G\frac{m_{1}m_{2}}{3.03}$

$\frac{9}{3.03} \cong 3$

$F_{g2}\approx3F_{g1}$

Compare your answer with the correct one above

Question

Which of the following changes would increase a satellite's orbital speed?

Answer

We know , and the force of gravity is:

$F_{grav}=\frac{GMm}{R^{2}}$

Also, the equation for uniform circular motion, such as a satellite in orbit is:

$a=\frac{v^{2}}{R}$

Set $F_{grav}=ma$ and substitute the acceleration due to circular motion into the equation. Solve for velocity.

$v = \sqrt{\frac{GM}{R}}$

This indicates that the only variables that affect the orbital speed are orbital radius and the mass of the Earth.

Compare your answer with the correct one above

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