Question 1

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

Accepted Answer

The orbital period of a satellite around a planet depends on the altitude of the orbit and the mass of the planet, not the mass of the satellite. Therefore, doubling the mass of the satellite does not change the orbital period.

Question 2

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

Accepted Answer

According to Kepler's Third Law, $T^2 \propto a^3$, where $T$ is the period of the orbit and $a$ is the semi-major axis of the ellipse. Since the distance from the center of the planet is the same in both cases, the semi-major axis is not changing, and therefore the period remains the same. The centripetal force keeping the satellite in orbit is given by $F = \frac{GMm}{r^2}$, where $r$ is the distance from the center of the planet, $M$ is the mass of the planet, and $m$ is the mass of the satellite. Therefore, the orbital speed of the satellite is proportional to $\sqrt{\frac{M}{r}}$. If $M$ is halved, the orbital speed becomes $\sqrt{\frac{M/2}{r}} = \frac{1}{\sqrt{2}}\sqrt{\frac{M}{r}}$, or $\frac{1}{\sqrt{2}}$ times the original speed. Therefore, the correct answer is D, because the orbital speed is halved.

Question 3

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)

Accepted Answer

The correct answer is (a) 2.4 kg × $10^{25}$ kg, (b) 1.2 kg ×$10 ^ { 31 }$ kg.The mass of the planet can be calculated using the formula for the acceleration due to gravity: g = GM/r^2, where g is the acceleration due to gravity, M is the mass of the planet, and r is the radius of the planet. The mass of the star can be calculated using the formula for the period of an orbit: T^2 = (4π^2/GM)r^3, where T is the period of the orbit, M is the mass of the star, and r is the distance from the planet to the star.

Question 4

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)

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 5

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)

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 6

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)

Accepted Answer

We can use the formula for the period of an orbit in terms of radius and mass:

$T=2\pi\sqrt{\frac{r^3}{GM}}$

where $T$ is the period, $r$ is the radius of the orbit, $M$ is the mass of the central body, and $G$ is the gravitational constant.

Plugging in the given values, we have:

$1.77	ext{ days} = 2\pi\sqrt{\frac{(422	imes10^6	ext{ m})^3}{(6.67	imes10^{-11}	ext{ N}\cdot	ext{m}^2/	ext{kg}^2)M}}$

Solving for $M$, we get:

$M = \frac{(422	imes10^6	ext{ m})^3}{(1.77	ext{ days}/2\pi)^2(6.67	imes10^{-11}	ext{ N}\cdot	ext{m}^2/	ext{kg}^2)}$

$M = 1.89	imes10^{27}$ kg

Therefore, the answer is D: 1.89 × 10^27 kg.

Question 7

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)

Accepted Answer

We can use Kepler's Third Law to solve for the mass of Jupiter. Kepler's Third Law states that the square of the period of a planet's orbit is proportional to the cube of the semi-major axis of its orbit.

Let T be the period of Europa's orbit and a be the semi-major axis of Europa's orbit. Then we have:

T^2 = (4π^2 / GM) * a^3

where G is the gravitational constant and M is the mass of Jupiter.

We can solve for M by rearranging this equation:

M = (4π^2 / G) * (a^3 / T^2)

We are given a = 1.34 × 10^9 m and T = 3.55 days = 3.55 * 24 * 60 * 60 = 306,720 seconds.

Plugging in these values and using G = 6.67 × 10^-11 N m^2/kg^2, we get:

M = (4π^2 / 6.67 × 10^-11) * ((1.34 × 10^9)^3 / (306,720)^2) = 1.89 × 10^27 kg

Therefore, the best choice is D, 1.89 × 10^27 kg.

Question 8

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 9

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)

Accepted Answer

The mass of the planet can be calculated using the following equation:

(T^2 * G * M) / (4π^2) = R^3

Where T is the orbital period, G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet plus the altitude of the satellite.

Substituting the given values, we get:

(89*60)^2 * 6.67 × 10^-11 * M / (4π^2) = (6.38 × 10^6 + 230000)^3

Simplifying, we get:

M = ((89*60)^2 * 6.67 × 10^-11 * (6.38 × 10^6 + 230000)^3) / (4π^2)

Calculating this expression using a calculator gives a value of approximately 6.0 × 10^24 kg, which is closest to option C. Therefore, the best choice is C.

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

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 12

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 13

$\begin{array} { | l | c | c | c | c | } \hline & \text { Mass } & \text { Radius } & \text { Orbital radius } & \text { Orbital period } \ \hline \text { Moon A } & 4.0 \times 10 ^ { 20 } \mathrm {~kg} & \text { unknown } & 2.0 \times 10 ^ { 8 } \mathrm {~m} & 4.0 \times 10 ^ { 6 } \mathrm {~s} \ \hline \text { Moon B } & 1.5 \times 10 ^ { 20 } \mathrm {~kg} & 2.0 \times 10 ^ { 5 } \mathrm {~m} & 3.0 \times 10 ^ { 8 } \mathrm {~m} & \text { unknown } \ \hline \end{array}$ 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)

Accepted Answer

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

Question 14

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)

Accepted Answer

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

Question 15

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)

Accepted Answer

The centripetal force required for circular orbit is provided by gravitational force: $ \frac{mv^2}{r} = \frac{GmM}{r^2} $. Solving for $ M $, the mass of the planet, gives $ M = \frac{v^2r}{G} $. Substituting $ v = 68 \, \mathrm{m/s} $, $ r = 410,000 \, \mathrm{m} $, and $ G = 6.67 \times 10^{-11} \, \mathrm{N \cdot m^2/kg^2} $, we find $ M \approx 2.8 \times 10^{19} \, \mathrm{kg} $.

Question 16

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)

Accepted Answer

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

Quiz 6: Gravity

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 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

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)

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)

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 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 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 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)

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)

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)

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)