Question 1

If the mass of the earth and all objects on it were suddenly doubled, but the size remained the same, the acceleration due to gravity at the surface would become

Accepted Answer

The acceleration due to gravity is directly proportional to the mass of the object and inversely proportional to the square of the distance between the centers of the two objects. Doubling the mass of the earth and all objects on it will double the gravitational force acting on objects on the surface, but the distance between the center of the earth and objects on its surface remains the same. Therefore, the acceleration due to gravity at the surface will also double.

Question 2

A satellite of mass m has an orbital period T when it is in a circular orbit of radius R around the earth. If the satellite instead had mass 4m, its orbital period would be

Accepted Answer

The orbital period of a satellite in a circular orbit around the Earth depends on the radius of the orbit and the mass of the Earth, but not on the mass of the satellite. Therefore, changing the mass of the satellite does not affect its orbital period.

Question 3

Two planets having equal masses are in circular orbit around a star. Planet A has a smaller orbital radius than planet B. Which statement is true?

Accepted Answer

Planet A, being closer to the star, moves faster in its orbit due to the stronger gravitational pull, thus having more kinetic energy. However, its potential energy is less because potential energy is more negative closer to the star, leading to a lower total mechanical energy compared to Planet B.

Question 4

The gravitational acceleration on a planet's surface is 16.0 m/s². What is the gravitational acceleration at an altitude of one planet diameter above the SURFACE of the planet?

Accepted Answer

The answer of The gravitational acceleration on a planet's surface...

Question 5

A satellite in a circular orbit of radius R around planet X has an orbital period T. If Planet X had one-fourth as much mass, the orbital period of this satellite in an orbit of the same radius would be

Accepted Answer

The answer of A satellite in a circular orbit of...

Question 6

An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 581 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. What is the ratio of the masses X/Y of the asteroids?

Accepted Answer

The gravitational force between the astronaut and asteroid X is equal to the gravitational force between the astronaut and asteroid Y. Therefore, the ratio of the masses of the asteroids is equal to the ratio of the squares of the distances between the astronaut and the asteroids.$$M_X/M_Y = (140 km/581 km)^2 = 0.0581$$

Question 7

A certain planet has an escape speed V. If another planet of the same size has twice the mass as the first planet, its escape speed will be

Accepted Answer

The answer of A certain planet has an escape speed...

Question 8

A satellite is orbiting the earth. If a payload of material is added until it doubles the satellite's mass, the earth's pull of gravity on this satellite will double but the satellite's orbit will not be affected.

Accepted Answer

The addition of payload will increase the mass of the satellite but will not affect its velocity. According to the law of conservation of angular momentum, the orbit of a satellite is determined solely by its velocity and the mass of the object it is orbiting around. Therefore, the orbit will not be affected. However, the earth's pull of gravity on the satellite will increase because the force of gravity is directly proportional to the mass of the object.

Question 9

The radius of the earth is R. At what distance above the earth's surface will the acceleration of gravity be 4.9 m/s²?

Accepted Answer

The acceleration of gravity is inversely proportional to the square of the distance from the center of the earth. So, if the acceleration of gravity is 4.9 m/s² at a distance of R from the center of the earth, then it will be 4.9 m/s² / (1.4)² = 2.5 m/s² at a distance of 1.4R from the center of the earth.

Question 10

A certain planet has an escape speed V. If another planet has twice size and twice the mass of the first planet, its escape speed will be

Accepted Answer

The answer of A certain planet has an escape speed...

Question 11

Planet Z-34 has a mass equal to 1/3 that of Earth, a radius equal to 1/3 that of Earth, and an axial spin rate 1/2 that of Earth. With g representing, as usual, the acceleration due to gravity on the surface of Earth, the acceleration due to gravity on the surface of Z-34 is

Accepted Answer

The acceleration due to gravity (g) on the surface of a planet is directly proportional to the mass of the planet (M) and inversely proportional to the square of the radius (R) of the planet, as given by the formula g = GM/R^2, where G is the gravitational constant. For planet Z-34, with a mass 1/3 that of Earth and a radius 1/3 that of Earth, the acceleration due to gravity on Z-34 would be (1/3)M / (1/3)^2R^2 = (1/3) / (1/9) = 3 times the acceleration due to gravity on Earth, or 3g. The axial spin rate does not affect the calculation of gravitational acceleration.

Question 12

Three identical very small 50-kg masses are held at the corners of an equilateral triangle, 0.30 m on each side. If one of the masses is released, what is its initial acceleration if the only forces acting on it are the gravitational forces due to the other two masses? (G = 6.67 × 10^-11N ∙ m²/kg²)

Accepted Answer

The answer of Three identical very small 50-kg masses are...

Question 13

Because the earth's orbit is slightly elliptical, the earth actually gets closer to the sun during part of the year. When the earth is closer to the sun its orbital speed is

Accepted Answer

According to Kepler's Second Law, a planet moves faster when it is closer to the sun and slower when it is farther away. Therefore, when the earth is closer to the sun, its orbital speed is greater than when it is farthest away from the sun.

Question 14

A very small round ball is located near a large solid sphere of uniform density. The force that the large sphere exerts on the ball

Accepted Answer

According to the shell theorem, the gravitational force exerted by a spherically symmetric object (such as a large solid sphere of uniform density) on a point object lying outside it is the same as if its entire mass were concentrated at its center. Therefore, the force that the large sphere exerts on the small ball is exactly the same as if all the mass of the sphere were concentrated at the center.

Question 15

The reason an astronaut in an earth satellite feels weightless is that

Accepted Answer

An astronaut in an earth satellite feels weightless because they are falling around the earth, but because the satellite is moving forward at the right speed, they are continuously missing the earth. This is sometimes referred to as being in a state of freefall, which is why they do not experience a sensation of weight.

Question 16

A small-sized 155-kg mass is located 1.50 m from a small-sized 275-kg mass, with both masses fixed in place. Where should you place a third small-sized mass so that the net gravitational force on it due to the original two masses is zero?

Accepted Answer

The answer of A small-sized 155-kg mass is located 1.50...

Question 17

By how many newtons does the weight of a 100-kg person change when he goes from sea level to an altitude of 5.0 km if we neglect the earth's rotational effects? (The mean radius of the Earth is 6.38 × 10⁶ m, G = 6.67 × 10^-11 N ∙ m²/kg².)

Accepted Answer

The answer of By how many newtons does the weight...

Question 18

A small planet having a radius of 1000 km exerts a gravitational force of 100 N on an object that is 500 km above its surface. If this object is moved 500 km farther from the planet, the gravitational force on it will be closest to

Accepted Answer

The force of gravity on an object can be calculated using the formula F = GMm/r^2, where G is the gravitational constant, M is the mass of the planet, m is the mass of the object, and r is the distance between the centers of the planet and the object.

Since we are given the radius of the planet and the force of gravity on the object when it is 500 km above the surface, we can solve for the mass of the planet using the following equation:

F = GMm/r^2

100 N = G(Mm)/(1000+500)^2

100 N = GMm/1.5^2 * 1000^2

M = (100 N * 1.5^2 * 1000^2) / (G * m)

Now we can use this mass value and the new distance (1000 + 500 + 500 = 2000 km) to calculate the new gravitational force:

F = GMm/r^2

F = G(Mm)/(2000 km)^2

F = (100 N * 1.5^2 * 1000^2) / (G * m * 2000^2)

F = 55.96 N, which is closest to answer choice C (56 N).

Question 19

A baseball is located at the surface of the earth. Which statements about it are correct? (There may be more than one correct choice.)

Accepted Answer

According to Newton's third law of motion, every action has an equal and opposite reaction. This means that the gravitational force the Earth exerts on the baseball is exactly equal in magnitude to the gravitational force the baseball exerts on the Earth, making option C correct.

Question 20

The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is 389 N. If he moves to a distance of 1.86 × 10⁴ km above the planet's surface, his weight changes to 24.31 N. What is the mass of planet X, if Speff's mass is 75.0 kg? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Accepted Answer

The gravitational force is given by F = G * (m1 * m2) / r^2. At the surface, F1 = 389 N, and at a distance above the surface, F2 = 24.31 N. Using the ratio of forces and distances, we can solve for the mass of the planet. The correct mass is 2.99 × 10^24 kg.

Quiz 12: Gravitation