Question 1

What would be the weight of a 59.1-kg astronaut on a planet with the same density as Earth and having twice Earth's radius?

Accepted Answer

The answer of What would be the weight of a...

Question 2

The mass of Pluto is 1.31 × 10²² kg and its radius is 1.15 × 10⁶ m.What is the acceleration of a freely-falling object at the surface of Pluto if it has no atmosphere? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Accepted Answer

The acceleration due to gravity on the surface of a celestial body is given by the formula $g = \frac{G \cdot M}{r^2}$, where $G$ is the gravitational constant, $M$ is the mass of the body, and $r$ is its radius. Substituting the given values for Pluto ($M = 1.31 \times 10^{22}$ kg, $r = 1.15 \times 10^{6}$ m, and $G = 6.67 \times 10^{-11}$ N·m²/kg²) into the formula yields an acceleration of approximately $0.661$ m/s².

Question 3

An astronaut goes out for a "space-walk" at a distance above the earth equal to the radius of the earth.What is her acceleration due to gravity at that point?

Accepted Answer

The answer of An astronaut goes out for a "space-walk"...

Question 4

Mithra is an unknown planet that has two airless moons,A and B,in circular orbits around it.The table summarizes the hypothetical data about these moons.If you dropped a laser at the surface of Moon B,at what rate would it accelerate toward the ground? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Accepted Answer

The answer of 
Mithra is an unknown planet that has...

Question 5

Planet Z-34 has a mass equal to one-third that of Earth and a radius equal to one-third that of Earth.With g representing,as usual,the acceleration due to gravity at the surface of Earth,the acceleration due to gravity at the surface of Z-34 is

Accepted Answer

The acceleration due to gravity is proportional to the mass and inversely proportional to the square of the radius. Since the mass and radius of Z-34 are both one-third that of Earth, the acceleration due to gravity at the surface of Z-34 would be (1/3)/(1/3)^2 * g = 3g. Therefore, the answer is choice B.

Question 6

A spaceship with a mass of 2.8 × 10⁶ kg is traveling toward two spherical asteroids,each with a mass of 5.0 × 10¹⁶ kg,that are 40 km apart center-to-center.Its path is perpendicular to the line joining the asteroids and is aimed at the midpoint of that line.What is the net gravitational force exerted by the asteroids on the spaceship when the spaceship is 30 km away from that midpoint? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Accepted Answer

The net gravitational force on the spaceship is the vector sum of the gravitational forces exerted by the two asteroids. The gravitational force exerted by each asteroid is given by:$$F = G \frac{m_1 m_2}{r^2}$$where:* G is the gravitational constant (6.67 × 10^-11 N ∙ m^2/kg^2)* m_1 is the mass of the asteroid (5.0 × 10^16 kg)* m_2 is the mass of the spaceship (2.8 × 10^6 kg)* r is the distance between the asteroid and the spaceship (30 km = 30,000 m)The net gravitational force is:$$F_net = F_1 + F_2 = G \frac{m_1 m_2}{r^2} + G \frac{m_1 m_2}{r^2} = 2G \frac{m_1 m_2}{r^2}$$Plugging in the values, we get:$$F_net = 2(6.67 × 10^-11 N ∙ m^2/kg^2) \frac{(5.0 × 10^16 kg)(2.8 × 10^6 kg)}{(30,000 m)^2} = 12,000 N$$Therefore, the correct answer is A.

Question 7

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 due to gravity at a distance $r$ from the center of the Earth (where $r > R$, $R$ being the Earth's radius) is given by $g' = \frac{G M}{r^2}$, where $G$ is the gravitational constant and $M$ is the mass of the Earth. At the surface of the Earth, the acceleration due to gravity $g = \frac{G M}{R^2} = 9.8 \, \text{m/s}^2$. To find the distance $r$ where the acceleration due to gravity is $4.9 \, \text{m/s}^2$, we set $g' = 4.9 \, \text{m/s}^2$ and solve for $r$:$$4.9 = \frac{G M}{r^2}$$Given that $g = 9.8 = \frac{G M}{R^2}$, we can substitute $9.8$ for $\frac{G M}{R^2}$ in the equation:$$4.9 = \frac{9.8 R^2}{r^2}$$Simplifying, we get:$$1 = \frac{9.8 R^2}{4.9 r^2}$$$$1 = \frac{2 R^2}{r^2}$$Taking the square root of both sides:$$1 = \frac{\sqrt{2} R}{r}$$$$r = \sqrt{2} R$$To find the distance above the Earth's surface, we subtract the Earth's radius $R$ from $r$:$$r - R = (\sqrt{2} - 1)R \approx 0.41 R$$Therefore, the acceleration of gravity will be $4.9 \, \text{m/s}^2$ at a distance of approximately $0.41 R$ above the Earth's surface.

Question 8

The mass of the Moon is 7.4 × 10²² kg,its radius is 1.74 × 10³ km,and it has no atmosphere.What is the acceleration due to gravity at the surface of the Moon? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Accepted Answer

The acceleration due to gravity on the surface of a celestial body can be calculated using the formula $g = \frac{G \cdot M}{r^2}$, where $G$ is the gravitational constant, $M$ is the mass of the body, and $r$ is its radius. Substituting the given values for the Moon ($G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2$, $M = 7.4 \times 10^{22} \, \text{kg}$, $r = 1.74 \times 10^6 \, \text{m}$), we find that the acceleration due to gravity on the Moon's surface is approximately $1.6 \, \text{m/s}^2$.

Question 9

At a distance of 14,000 km from the center of Planet Z-99,the acceleration due to gravity is 32 m/s².What is the acceleration due to gravity at a point 28,000 km from the center of this planet?

Accepted Answer

The acceleration due to gravity is inversely proportional to the square of the distance from the center of the planet. Doubling the distance (from 14,000 km to 28,000 km) decreases the acceleration by a factor of 4 (since 2^2 = 4), so 32 m/s^2 divided by 4 equals 8 m/s^2.

Question 10

At a given point above Earth's surface,the acceleration due to gravity is equal to What is the altitude of this point above Earth's surface? (G = 6.67 × 10^-11 N ∙ m²/kg²,M_earth = 5.97 × 10²⁴ kg,R_earth = 6.38 × 10⁶ m)

Accepted Answer

The answer of At a given point above Earth's surface,the...

Question 11

What would be the weight of a 59.1-kg astronaut on a planet twice as massive as Earth and having twice Earth's radius?

Accepted Answer

The gravitational force is proportional to the mass of the planet and inversely proportional to the square of the radius. Doubling the mass and radius results in a gravitational force of 1/2 that on Earth. The astronaut's weight on Earth is 580 N, so on the new planet, it would be 290 N.

Question 12

From what height above the surface of the earth should an object be dropped to initially experience an acceleration of 0.54g? (G = 6.67 × 10^-11 N ∙ m²/kg²,M_earth = 5.97 × 10²⁴ kg,R_earth = 6.38 × 10⁶ m)

Accepted Answer

The acceleration due to gravity at a height h above the surface of the Earth is given by:$$g' = \frac{GM_e}{(R_e + h)^2}$$where G is the gravitational constant, M_e is the mass of the Earth, and R_e is the radius of the Earth.We want to find the height h at which g' = 0.54g. Substituting the given values into the equation, we get:$$0.54g = \frac{GM_e}{(R_e + h)^2}$$$$0.54(9.81 m/s^2) = \frac{(6.67 × 10^{-11} N ∙ m^2/kg^2)(5.97 × 10^{24} kg)}{(6.38 × 10^6 m + h)^2}$$Solving for h, we get:$$h = 2.3 × 10^6 m = 2300 km$$Therefore, the correct answer is A) 2300 km.

Question 13

At their closest approach,Venus and Earth are 4.20 × 10¹⁰ m apart.The mass of Venus is 4.87 × 10²⁴ kg,the mass of Earth is 5.97 × 10²⁴ kg,and G = 6.67 × 10^-11 N ∙ m²/kg².What is the magnitude of the gravitational force exerted by Venus on Earth at that point?

Accepted Answer

The gravitational force can be calculated using Newton's law of universal gravitation: $F = \frac{G \cdot m_1 \cdot m_2}{r^2}$, where $G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2$, $m_1 = 4.87 \times 10^{24} \, \text{kg}$ (mass of Venus), $m_2 = 5.97 \times 10^{24} \, \text{kg}$ (mass of Earth), and $r = 4.20 \times 10^{10} \, \text{m}$ (distance between Venus and Earth). Plugging these values into the formula gives a force of approximately $1.10 \times 10^{18} \, \text{N}$.

Question 14

The earth has radius R.A satellite of mass 100 kg is in orbit at an altitude of 3R above the earth's surface.What is the satellite's weight at the altitude of its orbit?

Accepted Answer

The weight of an object in orbit can be calculated using the formula for gravitational force: $F = \frac{Gm_1m_2}{r^2}$, where $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses of the two objects, and $r$ is the distance between their centers. In this case, $r = 4R$ (since the satellite is 3R above the Earth's surface, the total distance from the Earth's center is the Earth's radius $R$ plus 3R). The gravitational force (weight) decreases with the square of the distance, so at 4R, the weight is $\frac{1}{16}$th of its weight on the surface (since $(4R)^2 = 16R^2$). If the satellite's weight on the surface is $980 N$ (using $g = 9.8 m/s^2$ for Earth's surface gravity), at 4R from the center, its weight would be $\frac{980 N}{16} = 61.25 N$, which rounds to $61 N$, making A the correct choice.

Question 15

An object weighs 432 N on the surface of the earth.The earth has radius R.If the object is raised to a height of 3R above the earth's surface,what is its weight?

Accepted Answer

The weight of an object above the surface of the earth can be calculated using the formula:

Weight = (mass x gravitational acceleration) / (distance from center of earth)^2

Since the mass of the object is unknown, it can be cancelled out by dividing the weight at the surface of the earth by the weight at the new height:

Weight ratio = Weight at 3R / Weight at surface
= [(mass x gravitational acceleration) / (4R^2)] / [(mass x gravitational acceleration) / R^2]
= 1 / 16

Therefore, Weight at 3R = Weight at surface x (1/16)
= 432 N x (1/16)
= 27.0 N

Hence, the answer is option C) 27.0 N.

Question 16

In another solar system,a planet has an airless moon Zygo that is 4.0 × 10⁵ m in diameter.Experiments reveal that a freely falling object at the surface of Zygo accelerates at 0.20 m/s².What is the mass of Zygo? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Accepted Answer

The gravitational acceleration formula is g = G * M / r^2. Solving for M gives M = g * r^2 / G. Substituting g = 0.20 m/s², r = 2.0 × 10^5 m, and G = 6.67 × 10^-11 N∙m²/kg², we find M ≈ 1.2 × 10^20 kg.

Question 17

By how many newtons does the weight of a 100-kg person decrease when he goes from sea level to mountain top at an altitude of 5000 m? The mean radius of the earth is 6.38 × 10⁶ m.

Accepted Answer

The weight decrease can be calculated using the formula for gravitational force: F = G * (m1 * m2) / r^2. The change in weight is due to the change in distance from the center of the Earth. At 5000 m, the distance is 6.38 × 10^6 m + 5000 m. The difference in gravitational force at sea level and at 5000 m gives the decrease in weight, which is approximately 1.5 N.

Question 18

What is the distance from the center of the Moon to the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal? The mass of Earth is 5.97 × 10²⁴ kg,the mass of the Moon is 7.35 × 10²² kg,the center-to-center distance between Earth and the Moon is 3.84 × 10⁸ m,and G = 6.67 × 10^-11 N ∙ m²/kg².

Accepted Answer

The distance from the center of the Moon to the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal is known as the barycenter. The location of the barycenter can be calculated using the formula:

r = R (m1 / (m1 + m2))

where r is the distance from the center of the Moon to the barycenter, R is the distance between the centers of Earth and Moon, m1 is the mass of Earth, and m2 is the mass of the Moon.

Plugging in the given values, we get:

r = (3.84 × 10^8 m) (5.97 × 10^24 kg / (5.97 × 10^24 kg + 7.35 × 10^22 kg))
r = 3.84 × 10^8 m (0.987)
r = 3.79 × 10^8 m

Therefore, the distance from the center of the Moon to the barycenter is approximately 3.79 × 10^8 m. However, we need to add the radius of the Moon (1.74 × 10^6 m) to this value to get the distance from the surface of the Moon to the barycenter:

3.79 × 10^8 m + 1.74 × 10^6 m = 3.84 × 10^8 m

Hence, the best choice is B.

Question 19

An astronaut drops a marble on the surface of the airless Planet Z-49 and observes that it takes 1.02 s for the marble to fall 2.00 m starting from rest.She also knows that the radius of Z-49 is 3.39 × 10⁶ m.From this information,what will she determine for the mass of Z-49? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Accepted Answer

Using the equation for the acceleration due to gravity,

g = GM/r^2

where g is the acceleration due to gravity, G is the gravitational constant, M is the mass of the planet, and r is the radius of the planet, we can solve for M:

M = gr^2/G

Using the given values, 
g = 2.00 m / (1.02 s)^2 = 1.942 m/s^2 
r = 3.39 × 10^16 m

Plugging in these values for g, r, and G, we get:

M = (1.942 m/s^2) x (3.39 × 10^16 m)^2 / (6.67 × 10^-11 N*m^2/kg^2) = 6.62 x 10^11 kg

Therefore, the answer is B: 6.62 × 10^11 kg.

Question 20

Three identical 50-kg balls are held at the corners of an equilateral triangle,30 cm on each side.If one of the balls is released,what is the magnitude of its initial acceleration if the only forces acting on it are the gravitational forces due to the other two masses? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Accepted Answer

The answer of Three identical 50-kg balls are held at...

Quiz 12: Gravity

What would be the weight of a 59.1-kg astronaut on a planet with the same density as Earth and having twice Earth's radius?

The mass of Pluto is 1.31 × 1022 kg and its radius is 1.15 × 106 m.What is the acceleration of a freely-falling object at the surface of Pluto if it has no atmosphere? (G = 6.67 × 10-11 N ∙ m2/kg2)

An astronaut goes out for a "space-walk" at a distance above the earth equal to the radius of the earth.What is her acceleration due to gravity at that point?

Mithra is an unknown planet that has two airless moons,A and B,in circular orbits around it.The table summarizes the hypothetical data about these moons.If you dropped a laser at the surface of Moon B,at what rate would it accelerate toward the ground? (G = 6.67 × 10-11 N ∙ m2/kg2)

Planet Z-34 has a mass equal to one-third that of Earth and a radius equal to one-third that of Earth.With g representing,as usual,the acceleration due to gravity at the surface of Earth,the acceleration due to gravity at the surface of Z-34 is

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?

The mass of the Moon is 7.4 × 1022 kg,its radius is 1.74 × 103 km,and it has no atmosphere.What is the acceleration due to gravity at the surface of the Moon? (G = 6.67 × 10-11 N ∙ m2/kg2)

At a distance of 14,000 km from the center of Planet Z-99,the acceleration due to gravity is 32 m/s2.What is the acceleration due to gravity at a point 28,000 km from the center of this planet?

At a given point above Earth's surface,the acceleration due to gravity is equal to What is the altitude of this point above Earth's surface? (G = 6.67 × 10-11 N ∙ m2/kg2,Mearth = 5.97 × 1024 kg,Rearth = 6.38 × 106 m)