Question 1

In another solar system, a planet has a moon that is 4.0 × 10⁵ m in diameter. Measurements reveal that this moon takes 3.0 x 10⁵ s to make each orbit of diameter 1.8 × 10⁸ m. What is the mass of the planet? (G = 6.67 × 10^-11 N ∙ m²/kg²) A)1.2 × 10²⁴ kg B)1.7 × 10²⁴ kg C)2.4 × 10²⁴ kg D)3.4 × 10²⁴ kg E)4.8 × 10²⁴ kg

Accepted Answer

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Question 2

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 $T$ can be calculated using the formula for the orbital period of a satellite revolving around Earth, $T = 2\pi\sqrt{\frac{r^3}{GM}}$, where $r$ is the radius of the orbit, $G$ is the gravitational constant, and $M$ is the mass of Earth. Given that the satellite revolves around Earth 5 times a day, we can convert this to seconds to find $T$. Since there are 86400 seconds in a day, $T = \frac{86400}{5}$ seconds for each orbit. Rearranging the formula to solve for $r$ and substituting the given values for $G$ and $M_{earth}$, we find that the radius $r$ that satisfies the condition for 5 orbits per day is closest to the value given in option A, $1.44 \times 10^7 m$. This calculation involves algebraic manipulation and substitution into the rearranged orbital period formula.

Question 3

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

Accepted Answer

The answer of Find the orbital speed of an ice...

Question 4

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

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Question 5

A satellite that is in a circular orbit 230 km above the surface of the planet Zeeman-474 has an orbital period of 89 min. The radius of Zeeman-474 is 6.38 × 10⁶ m. What is the mass of this planet? (G = 6.67 × 10^-11 N ∙ m²/kg²) A)5.0 × 10²⁴ kg B)5.5 × 10²⁴ kg C)6.0 × 10²⁴ kg D)6.5 × 10²⁴ kg

Accepted Answer

We can use the formula for orbital period of a circular orbit:
T = 2π√(r³/GM)

where T is the orbital period, r is the radius of the orbit, G is the gravitational constant, M is the mass of the planet.

Rearranging this formula, we get:

M = (4π²r³)/(GT²)

Plugging in the given values, we get:

M = (4π² × (6.38 × 10^16)³)/(6.67 × 10^-11 × (89 × 60)^2)

M = 6.0 × 10^24 kg

Therefore, the mass of the planet Zeeman-474 is 6.0 × 10^24 kg, which is answer choice C.

Question 6

Europa, a satellite of Jupiter, has an orbital diameter of 1.34 × 10⁹ m and a period of 3.55 days. What is the mass of Jupiter? (G = 6.67 × 10^-11 N ∙ m²/kg²) A)1.53 × 10²⁷ kg B)1.65 × 10²⁷ kg C)1.07 × 10²⁷ kg D)1.89 × 10²⁷ kg E)3.08 × 10²⁷ kg

Accepted Answer

The answer of Europa, a satellite of Jupiter, has an...

Question 7

You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8 × 10⁷ m. You have previously determined that the planet orbits 2.9 × 10¹¹ m from its star with a period of 402 earth days. Once on the surface you find that the acceleration due to gravity is 19.5 m/s². What are the masses of (a)the planet and (b)the star? (G = 6.67 × 10^-11 N ? m²/kg²) A)(a)2.4 kg × $10^{25}$ kg, (b)1.2 kg × $10 ^ { 31 }$ kg B)(a)4.3 kg × $10^{25}$ kg, (b)1.2 kg × $10 ^ { 31 }$ kg C)(a)2.4 kg × $10^{25}$ kg, (b)7.1 kg × $10 ^ { 30 }$ kg D)(a)4.3 kg × $10^{25}$ kg, (b)7.1 kg × $10 ^ { 30 }$ kg

Accepted Answer

The correct answer is (a) 2.4 kg × $10^{25}$ kg and (b) 1.2 kg × $10 ^ { 31 }$ kg.The mass of the planet can be calculated using the formula:$$M_p = \frac{4\pi^2}{G} \times \frac{r^3}{T^2}$$where:* $M_p$ is the mass of the planet* $G$ is the gravitational constant* $r$ is the distance from the planet to the star* $T$ is the period of the planet's orbitPlugging in the given values, we get:$$M_p = \frac{4\pi^2}{6.67 \times 10^{-11}} \times \frac{(2.9 \times 10^{11})^3}{(402 \times 86400)^2} = 2.4 \times 10^{25} \text{ kg}$$The mass of the star can be calculated using the formula:$$M_s = \frac{a_g r^2}{G}$$where:* $M_s$ is the mass of the star* $a_g$ is the acceleration due to gravity on the planet's surface* $r$ is the radius of the planetPlugging in the given values, we get:$$M_s = \frac{19.5 \times (1.8 \times 10^7)^2}{6.67 \times 10^{-11}} = 1.2 \times 10^{31} \text{ kg}$$

Question 8

Asteroid Ida was photographed by the Galileo spacecraft in 1993, and the photograph revealed that the asteroid has a small moon, which has been named Dactyl. From the dimensions of Ida and its general features, one can estimate the mass of Ida to be 4.5 × 10¹⁶ kg, and the distance between Dactyl and Ida is approximately 90 km. Assuming a circular orbit, what would be the orbital speed of Dactyl? (G = 6.67 × 10^-11 N ∙ m²/kg²) A)5.8 m/s B)11 m/s C)2.3 m/s D)2.9 m/s E)30 m/s

Accepted Answer

The orbital speed $v$ of Dactyl can be found using the formula for gravitational force and centripetal force: $v = \sqrt{\frac{GM}{r}}$, where $G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2$, $M = 4.5 \times 10^{16} \, \text{kg}$ is the mass of Ida, and $r = 90 \, \text{km} = 90,000 \, \text{m}$ is the distance between Dactyl and Ida. Plugging in the values, we get $v = \sqrt{\frac{6.67 \times 10^{-11} \times 4.5 \times 10^{16}}{90,000}} \approx 5.8 \, \text{m/s}$.

Question 9

The innermost satellite of Jupiter orbits the planet with a radius of 422 × 10³ km and a period of 1.77 days. What is the mass of Jupiter? (G = 6.67 × 10^-11 N ∙ m²/kg²) A)1.33 × 10²⁷ kg B)1.50 × 10²⁷ kg C)1.72 × 10²⁷ kg D)1.89 × 10²⁷ kg E)3.08 × 10²⁷ kg

Accepted Answer

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Question 10

A large telescope of mass 8410 kg is in a circular orbit around the earth, making one revolution every 927 minutes. What is the magnitude of the gravitational force exerted on the satellite by the earth? (G = 6.67 × 10^-11 N ? m²/kg², M_earth = 6.0 × 10²⁴ kg)

Accepted Answer

The answer of A large telescope of mass 8410 kg...

Question 11

A satellite of mass 500 kg orbits the earth with a period of 6,000 s. The earth has a mass of 5.97 × 10²⁴kg, a radius of 6.38 × 10⁶ m, and G = 6.67 × 10^-11 N ? m²/kg². (a)Calculate the magnitude of the earth's gravitational force on the satellite. (b)Determine the altitude of the satellite above the earth's surface.

Accepted Answer

The answer of A satellite of mass 500 kg orbits...

Question 12

The International Space Station is orbiting at an altitude of about 370 km above the earth's surface. The mass of the earth is 5.97 × 10²⁴ kg, the radius of the earth is 6.38 × 10⁶ m, and G = 6.67 × 10^-11 N ? m²/kg². Assume a circular orbit. (a)What is the period of the International Space Station's orbit? (b)What is the speed of the International Space Station in its orbit?

Accepted Answer

The answer of The International Space Station is orbiting at...

Question 13

A satellite of mass M takes time T to orbit a planet. If the satellite had twice as much mass, the time for it to orbit the planet at the same altitude would be
A)4T.
B)2T.
C)T.
D)T/2.
E)T/4.

Accepted Answer

According to Kepler's Third Law, the time period of an orbiting body is proportional to the square root of the cube of its semimajor axis. The mass of the satellite does not directly affect its distance from the planet or its semimajor axis, so there is no change in the time period. Therefore, the time for the satellite with twice the mass to orbit the planet at the same altitude would be the same as the original time period T.

Question 14

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 gravitational force provides the necessary centripetal force for the orbit. Using the formula $v^2 = \frac{GM}{r}$, where $v = 68 \, \text{m/s}$, $r = 410,000 \, \text{m}$, and $G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2$, solving for $M$ gives approximately $2.8 \times 10^{19} \, \text{kg}$.

Question 15

A satellite having orbital speed V orbits a planet of mass M. If the planet had half as much mass, the orbital speed of the satellite at the same distance from the center would be
A)V $\sqrt { 2 }$ .
B)2V.
C)V.
D)V/ $\sqrt { 2 }$ .
E)V/2.

Accepted Answer

The orbital speed $ V $ of a satellite is given by $ V = \sqrt{\frac{GM}{r}} $. If the mass $ M $ of the planet is halved, the new speed $ V' $ becomes $ V' = \sqrt{\frac{G(M/2)}{r}} = \frac{V}{\sqrt{2}} $.

Question 16

$$\begin{array} { | l | c | c | c | c | } \hline & \text { Mass } & \text { Radius } & \text { Orbital radius } & \text { Orbital period } \ \hline \text { Moon A } & 4.0 \times 10 ^ { 20 } \mathrm {~kg} & \text { unknown } & 2.0 \times 10 ^ { 8 } \mathrm {~m} & 4.0 \times 10 ^ { 6 } \mathrm {~s} \ \hline \text { Moon B } & 1.5 \times 10 ^ { 20 } \mathrm {~kg} & 2.0 \times 10 ^ { 5 } \mathrm {~m} & 3.0 \times 10 ^ { 8 } \mathrm {~m} & \text { unknown } \ \hline \end{array}$$ Mithra is an unknown planet that has two moons, A and B, in circular orbits around it. The table summarizes the hypothetical data about these moons. What is the mass of Mithra? (G = 6.67 × 10^-11 N ? m²/kg²) A)1 × 10²² kg B)3 × 10²² kg C)1 × 10²³ kg D)3 × 10²³ kg E)1 × 10²⁴ kg

Accepted Answer

The answer of \[\begin{array} { | l | c |...

Quiz 6: Gravity

In another solar system, a planet has a moon that is 4.0 × 105 m in diameter. Measurements reveal that this moon takes 3.0 x 105 s to make each orbit of diameter 1.8 × 108 m. What is the mass of the planet? (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, Mearth = 5.97 × 1024 kg)

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 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)

A satellite that is in a circular orbit 230 km above the surface of the planet Zeeman-474 has an orbital period of 89 min. The radius of Zeeman-474 is 6.38 × 106 m. What is the mass of this planet? (G = 6.67 × 10-11 N ∙ m2/kg2)

Europa, a satellite of Jupiter, has an orbital diameter of 1.34 × 109 m and a period of 3.55 days. What is the mass of Jupiter? (G = 6.67 × 10-11 N ∙ m2/kg2)

The innermost satellite of Jupiter orbits the planet with a radius of 422 × 103 km and a period of 1.77 days. What is the mass of Jupiter? (G = 6.67 × 10-11 N ∙ m2/kg2)

A large telescope of mass 8410 kg is in a circular orbit around the earth, making one revolution every 927 minutes. What is the magnitude of the gravitational force exerted on the satellite by the earth? (G = 6.67 × 10-11 N ? m2/kg2, Mearth = 6.0 × 1024 kg)

A satellite of mass M takes time T to orbit a planet. If the satellite had twice as much mass, the time for it to orbit the planet at the same altitude would be

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)

A satellite having orbital speed V orbits a planet of mass M. If the planet had half as much mass, the orbital speed of the satellite at the same distance from the center would be

In another solar system, a planet has a moon that is 4.0 × 10⁵ m in diameter. Measurements reveal that this moon takes 3.0 x 10⁵ s to make each orbit of diameter 1.8 × 10⁸ m. What is the mass of the planet? (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)

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 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)

A satellite that is in a circular orbit 230 km above the surface of the planet Zeeman-474 has an orbital period of 89 min. The radius of Zeeman-474 is 6.38 × 10⁶ m. What is the mass of this planet? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Europa, a satellite of Jupiter, has an orbital diameter of 1.34 × 10⁹ m and a period of 3.55 days. What is the mass of Jupiter? (G = 6.67 × 10^-11 N ∙ m²/kg²)

The innermost satellite of Jupiter orbits the planet with a radius of 422 × 10³ km and a period of 1.77 days. What is the mass of Jupiter? (G = 6.67 × 10^-11 N ∙ m²/kg²)

A large telescope of mass 8410 kg is in a circular orbit around the earth, making one revolution every 927 minutes. What is the magnitude of the gravitational force exerted on the satellite by the earth? (G = 6.67 × 10^-11 N ? m²/kg², M_earth = 6.0 × 10²⁴ kg)

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)