Question 1

Suppose NASA wants a satellite to revolve around Earth 5 times a day.What should be the radius of its orbit if we neglect the presence of the Moon? (G = 6.67 × 10^-11 N ∙ m²/kg²,M_earth = 5.97 × 10²⁴ kg) A)1.44 × 10⁷ m B)0.69 × 10⁷ m C)7.22 × 10⁷ m D)2.11 × 10⁷ m

Accepted Answer

The orbital period $T$ for an object orbiting the Earth can be found using the formula derived from Newton's law of universal gravitation and circular motion principles: $T = 2\pi\sqrt{\frac{r^3}{GM}}$, where $r$ is the radius of the orbit, $G$ is the gravitational constant, and $M$ is the mass of the Earth. Given that the satellite revolves around Earth 5 times a day, we need to find the orbital period in seconds: $T = \frac{24 \times 3600}{5}$ seconds. Rearranging the formula to solve for $r$ and substituting the given values for $G$ and $M_{earth}$, we can solve for $r$. The correct radius that satisfies this condition is $1.44 \times 10^7 m$.

Question 2

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) A)2.4 × 10¹⁹ kg B)4.8 × 10¹⁹ kg C)1.2 × 10²⁰ kg D)2.4 × 10²⁰ kg E)4.8 × 10²⁰ kg

Accepted Answer

The answer of In another solar system,a planet has an...

Question 3

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) A)3.30 × $10^{23}$ kg. B)6.62 × $10^{23}$ kg. C)4.62 × $10^{23}$ kg. D)8.09 × $10^{23}$ kg. E)9.95 × $10^{23}$ kg.

Accepted Answer

The force of gravity acting on the falling marble is given by F = mg, where m is the mass of the planet and g is the acceleration due to gravity. Using the formula for the period of a satellite in circular orbit, we know that g = GM/r^2, where M is the mass of the planet, r is the radius of the planet, and G is the gravitational constant.

Thus, we have:

F = mg = (GM/r^2)m

Solving for M, we get:

M = (Fr^2)/Gm

Substituting the given values, we get:

M = [(2.00 kg)(3.39 × 10^16 m)^2]/[(6.67 × 10^-11 N ∙ m^2/kg^2)(1.02 s)^2]

M = 6.62 × 10^23 kg

Therefore, the answer is B.

Question 4

The captain of a spaceship orbiting planet X discovers that to remain in orbit at $410 \mathrm {~km}$ from the planet's center,she needs to maintain a speed of $68 \mathrm {~m} / \mathrm { s }$ What is the mass of planet X? (G = 6.67 × 10^-11 N ∙ m²/kg²) A)2.8 × 10¹⁹ kg B)4.2 × 10¹⁷ kg C)2.8 × 10¹⁶ kg D)4.2 × 10¹⁴ kg

Accepted Answer

The mass of planet X can be calculated using the formula for gravitational force $F = \frac{Gm_1m_2}{r^2}$ and the formula for centripetal force $F = \frac{mv^2}{r}$ required to maintain an orbit. Setting these equal to each other because the gravitational force provides the centripetal force needed for orbit, we get $\frac{Gm_1m_2}{r^2} = \frac{mv^2}{r}$. Simplifying and solving for $m_1$ (the mass of the planet), we get $m_1 = \frac{rv^2}{G}$. Plugging in the given values ($r = 410 \, \text{m}$, $v = 68 \, \text{m/s}$, $G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2$), we find the mass of planet X to be approximately $2.8 \times 10^{19} \, \text{kg}$.

Question 5

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 force that Earth exerts on the satellite at the altitude of its orbit?
A)61 N
B)110 N
C)9000 N
D)16,000 N

Accepted Answer

The gravitational force $F$ between the Earth and the satellite is given by $F = G \frac{Mm}{r^2}$, where $G$ is the gravitational constant, $M$ is the mass of the Earth, $m$ is the mass of the satellite, and $r$ is the distance from the center of the Earth to the satellite. At an altitude of $3R$, the distance $r = 4R$. Substituting the values and simplifying, we find that the force is closest to 61 N.

Question 6

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. A)0.60 N B)1.5 N C)2.6 N D)3.6 N E)9.8 N

Accepted Answer

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

Question 7

A spherically symmetric planet has four times the earth's mass and twice its radius.If a jar of peanut butter weighs 12 N on the surface of the Earth,how much would it weigh on the surface of this planet?
A)3.0 N
B)6.0 N
C)12 N
D)24 N
E)36 N

Accepted Answer

The answer of A spherically symmetric planet has four times...

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 free-fall acceleration at the surface of the Moon? (G = 6.67 × 10^-11 N ∙ m²/kg²) A)2.8 × 10⁶ m/s² B)9.8 m/s² C)4.9 m/s² D)1.6 m/s² E)0.80 m/s²

Accepted Answer

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

Question 9

$$\begin{array} { | l | c | c | c | c | } \hline & \text { Mass } & \text { Radius } & \text { Orbital radius } & \text { Orbital period } \ \hline \text { Moon A } & 4.0 \times 10 ^ { 20 } \mathrm {~kg} & \text { unknown } & 2.0 \times 10 ^ { 8 } \mathrm {~m} & 4.0 \times 10 ^ { 6 } \mathrm {~s} \ \hline \text { Moon B } & 1.5 \times 10 ^ { 2 } 0 \mathrm {~kg} & 2.0 \times 10 ^ { 5 } \mathrm {~m} & 3.0 \times 10 ^ { 8 } \mathrm {~m} & \text { unknown } \ \hline \end{array}$$ 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) A)0.10 m/s² B)0.15 m/s² C)0.20 m/s² D)0.25 m/s² E)0.30 m/s²

Accepted Answer

We can use Newton's law of gravitation to determine the acceleration due to gravity experienced by a laser at the surface of Moon B. 
From the table, we know that Moon B has a mass of $1.5 	imes 10^{20}$ kg and a radius of $2.0 	imes 10^5$ m. The distance between the center of Moon B and its orbital radius is given as $3.0 	imes 10^8$ m. 
Using Newton's law of gravitation, we have:
$$F = G \frac{mM}{r^2}$$
where $F$ is the force of gravity, $G$ is the gravitational constant, $m$ is the mass of the object (in this case, the laser), $M$ is the mass of Moon B, and $r$ is the radius of Moon B. 
Solving for $F$, we get:
$$F = G \frac{mM}{r^2} = (6.67 	imes 10^{-11} \mathrm{~N \cdot m^2/kg^2})(\frac{m}{4 	imes 10^{20} \mathrm{~kg}})(1.5 	imes 10^{20} \mathrm{~kg}) (\frac{1}{2 	imes 10^5 \mathrm{~m}})^2 = 1.25 	imes 10^{-6} \mathrm{~N}$$
The force of gravity is equal to the weight of the laser, so we can use $F = ma$ to find the acceleration $a$:
$$a = \frac{F}{m} = \frac{1.25 	imes 10^{-6} \mathrm{~N}}{m}$$
Since we want the acceleration in terms of meters per second squared, we need to convert the mass of the laser from kilograms to grams:
$$a = \frac{1.25 	imes 10^{-6} \mathrm{~N}}{m} = \frac{1.25 	imes 10^{-3} \mathrm{~N}}{1000m} = 1.25 	imes 10^{-6} \mathrm{~m/s^2}$$
Therefore, the laser would accelerate toward the ground at a rate of 0.25 m/s$^2$. The best choice is D.

Question 10

A satellite orbits the Earth once every 6.0 hours in a circle.What are the magnitude and direction of the acceleration of the satellite? (G = 6.67 × 10^-11 N ∙ m²/kg²,M_earth = 5.97 × 10²⁴ kg)

Accepted Answer

The answer of A satellite orbits the Earth once every...

Question 11

At a distance of 14,000 km from the center of Planet Z-99,the free-fall acceleration is 32 m/s².What is the free-fall acceleration at a point 28,000 km from the center of this planet? A)8.0 m/s² B)16 m/s² C)128 m/s² D)4.0 m/s² E)2.0 m/s²

Accepted Answer

The free-fall acceleration 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 12

A 2.10-kg hammer is transported to the Moon.The radius of the Moon is 1.74 × 10⁶ m,its mass is 7.35 × 10²² kg,and G = 6.67 × 10^-11 N ∙ m²/kg².What is the gravitational force exerted on the hammer on Earth and on the Moon?

Accepted Answer

The answer of A 2.10-kg hammer is transported to the...

Question 13

An astronaut goes out for a "space-walk" at a distance above the earth equal to the radius of the earth.What is her free-fall acceleration at that point?
A)zero
B)g
C)g/2
D)g/4
E)g/ $\sqrt { 2 }$

Accepted Answer

The free-fall acceleration at a distance equal to the radius of the Earth above the Earth's surface is given by the formula $ g' = \frac{g}{(2)^2} = \frac{g}{4} $, where g is the acceleration due to gravity on the Earth's surface.

Question 14

Earth exerts a gravitational force of 432 N on an object on its surface.Earth has radius R.If the object is raised to a height of 3R above Earth's surface,what is the gravitational force that Earth will exert on it?
A)108 N
B)48.0 N
C)27.0 N
D)305 N
E)144 N

Accepted Answer

The gravitational force exerted by the Earth on an object of mass 'm' at the surface of the Earth is given by:
F = G (Mm/R^2)
where G is the universal gravitational constant, M is the mass of the Earth, m is the mass of the object, and R is the radius of the Earth.

Given, F = 432 N, and let R' be the new distance between the center of the Earth and the object (3R).

So, the new gravitational force,
F' = G (Mm/R'^2)

Dividing F' by F,
F'/F = (R/R')^2

F' = F * (R/R')^2

Substituting the given values, we get
F' = 432 * ((1/3)^2)
F' = 432 * (1/9)
F' = 48 N
Therefore, the gravitational force that Earth will exert on the object when it is raised to a height of 3R above Earth's surface is 48 N, which is choice C).

Question 15

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 free-fall acceleration at the surface of Earth,the free-fall acceleration at the surface of Z-34 is
A)g/3.
B)3g.
C)6g.
D)g/9.
E)9g.

Accepted Answer

The equation for free-fall acceleration (g) at the surface of a planet is given by:

g = (G*M) / R^2

where G is the gravitational constant, M is the planet's mass, and R is the planet's radius.

For planet Z-34, we know that its mass is one-third that of Earth (M_Z-34 = (1/3)*M_Earth) and its radius is also one-third that of Earth (R_Z-34 = (1/3)*R_Earth).

Substituting these values into the equation for g, we get:

g_Z-34 = (G*((1/3)*M_Earth)) / ((1/3)*R_Earth)^2

g_Z-34 = (G*(1/3)*M_Earth) / ((1/9)*R_Earth^2)

g_Z-34 = (1/3)*(G*M_Earth) / R_Earth^2

g_Z-34 = (1/3)*g_Earth

Therefore, the free-fall acceleration at the surface of Z-34 is one-third that of Earth's (g_Z-34 = (1/3)*g_Earth) which means it is three times larger than Earth's (3g_Earth). Hence, the best choice is B.

Question 16

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) A)0.661 m/s² B)9.81 m/s² C)1.62 m/s² D)3.72 m/s² E)0.140 m/s²

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 ($G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2$, $M = 1.31 \times 10^{22} \, \text{kg}$, $r = 1.15 \times 10^{6} \, \text{m}$), the calculated acceleration is approximately $0.661 \, \text{m/s}^2$.

Question 17

What would be the gravitational force that a planet twice as massive as Earth and having twice Earth's radius exerts on a 59.1-kg astronaut?
A)580 N
B)290 N
C)1160 N
D)118 N
E)1200 N

Accepted Answer

The gravitational force is calculated using the formula F = G * (m1 * m2) / r^2. If the planet is twice as massive as Earth and has twice the radius, the force is (2 * M * m) / (2 * R)^2 = (2 * M * m) / (4 * R^2) = (1/2) * (M * m) / R^2. This means the force is half of what it would be on Earth. On Earth, the force is 580 N, so on this planet, it would be 290 N.

Question 18

The radius of the earth is R.At what distance above the earth's surface will the free-fall acceleration be 4.9 m/s²? A)0.41 R B)0.50 R C)1.00 R D)1.41 R E)0.25 R

Accepted Answer

The free-fall acceleration at a distance r from the center of the Earth is given by g' = GM/r^2, where G is the gravitational constant and M is the mass of the Earth. At the surface of the Earth, r = R and g' = g = 9.8 m/s^2. If g' = 4.9 m/s^2, then r = sqrt(GM/g') = sqrt(GM/4.9) = 1.41R. Therefore, the distance above the Earth's surface is 1.41R - R = 0.41R.

Question 19

What would be the gravitational force that a planet with the same density as Earth and having twice Earth's radius exerts on a 59.1-kg astronaut?
A)580 N
B)290 N
C)1160 N
D)2320 N
E)1200 N

Accepted Answer

We can use the formula for gravitational force:

F = G * (m1*m2)/r^2

Where:
 G = gravitational constant (6.67 x 10^-11 N*m^2/kg^2)
 m1 = mass of planet
 m2 = mass of astronaut
 r = distance between planet's center and astronaut's center

First, let's find the mass of the planet:

density = mass/volume
4/3 * π * r^3 * density = mass
4/3 * π * (2*6371 km)^3 * 5.5 g/cm^3 = 1.08 x 10^25 kg

Now we can plug in the values:

F = G * (m1*m2)/r^2
F = 6.67 x 10^-11 * (1.08 x 10^25 * 59.1) / (2*6371 km)^2
F = 1160 N

Therefore, the gravitational force that the planet exerts on the astronaut is 1160 N.

Question 20

Find the orbital speed of an ice cube in the rings of Saturn.The mass of Saturn is 5.68 x 10²⁶ kg,and use an orbital radius of 1.00 x 10⁵ km.(G = 6.67 × 10^-11 N ∙ m²/kg²) A)19.5 km/s B)27.5 km/s C)13.8 km/s D)9.40 km/s

Accepted Answer

The orbital speed of an object is given by:$$v = \sqrt{\frac{GM}{r}}$$where:* v is the orbital speed in m/s* G is the gravitational constant in N ∙ m^2/kg^2* M is the mass of the object in kg* r is the orbital radius in mPlugging in the given values, we get:$$v = \sqrt{\frac{(6.67 × 10^{-11} N ∙ m^2/kg^2)(5.68 × 10^{26} kg)}{1.00 × 10^5 km}}$$$$v = 19.5 km/s$$Therefore, the correct answer is A.

Quiz 5: Circular Motion

Suppose NASA wants a satellite to revolve around Earth 5 times a day.What should be the radius of its orbit if we neglect the presence of the Moon? (G = 6.67 × 10-11 N ∙ m2/kg2,Mearth = 5.97 × 1024 kg)

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)

The captain of a spaceship orbiting planet X discovers that to remain in orbit at 410 km410 \mathrm {~km}410 km from the planet's center,she needs to maintain a speed of 68 m/s68 \mathrm {~m} / \mathrm { s }68 m/s What is the mass of planet X? (G = 6.67 × 10-11 N ∙ m2/kg2)

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 force that Earth exerts on the satellite at the altitude of its orbit?

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.

A spherically symmetric planet has four times the earth's mass and twice its radius.If a jar of peanut butter weighs 12 N on the surface of the Earth,how much would it weigh on the surface of this planet?

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 free-fall acceleration at the surface of the Moon? (G = 6.67 × 10-11 N ∙ m2/kg2)

A satellite orbits the Earth once every 6.0 hours in a circle.What are the magnitude and direction of the acceleration of the satellite? (G = 6.67 × 10-11 N ∙ m2/kg2,Mearth = 5.97 × 1024 kg)

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

A 2.10-kg hammer is transported to the Moon.The radius of the Moon is 1.74 × 106 m,its mass is 7.35 × 1022 kg,and G = 6.67 × 10-11 N ∙ m2/kg2.What is the gravitational force exerted on the hammer on Earth and on the Moon?

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

Earth exerts a gravitational force of 432 N on an object on its surface.Earth has radius R.If the object is raised to a height of 3R above Earth's surface,what is the gravitational force that Earth will exert on it?

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 free-fall acceleration at the surface of Earth,the free-fall acceleration at the surface of Z-34 is

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)

What would be the gravitational force that a planet twice as massive as Earth and having twice Earth's radius exerts on a 59.1-kg astronaut?

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

What would be the gravitational force that a planet with the same density as Earth and having twice Earth's radius exerts on a 59.1-kg astronaut?

Find the orbital speed of an ice cube in the rings of Saturn.The mass of Saturn is 5.68 x 1026 kg,and use an orbital radius of 1.00 x 105 km.(G = 6.67 × 10-11 N ∙ m2/kg2)

Suppose NASA wants a satellite to revolve around Earth 5 times a day.What should be the radius of its orbit if we neglect the presence of the Moon? (G = 6.67 × 10^-11 N ∙ m²/kg²,M_earth = 5.97 × 10²⁴ kg)

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)

The captain of a spaceship orbiting planet X discovers that to remain in orbit at $410 \mathrm {~km}$ from the planet's center,she needs to maintain a speed of $68 \mathrm {~m} / \mathrm { s }$ What is the mass of planet X? (G = 6.67 × 10^-11 N ∙ m²/kg²)

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.

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 free-fall acceleration at the surface of the Moon? (G = 6.67 × 10^-11 N ∙ m²/kg²)

A satellite orbits the Earth once every 6.0 hours in a circle.What are the magnitude and direction of the acceleration of the satellite? (G = 6.67 × 10^-11 N ∙ m²/kg²,M_earth = 5.97 × 10²⁴ kg)

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

A 2.10-kg hammer is transported to the Moon.The radius of the Moon is 1.74 × 10⁶ m,its mass is 7.35 × 10²² kg,and G = 6.67 × 10^-11 N ∙ m²/kg².What is the gravitational force exerted on the hammer on Earth and on the Moon?

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)

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

Find the orbital speed of an ice cube in the rings of Saturn.The mass of Saturn is 5.68 x 10²⁶ kg,and use an orbital radius of 1.00 x 10⁵ km.(G = 6.67 × 10^-11 N ∙ m²/kg²)