Question 1

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

From the conservation of momentum, we have:

mv = m'v', where m is the mass of the planet, m' is the new mass of the planet (in this case, M/2), v is the initial orbital speed, and v' is the new orbital speed.

We can simplify this equation to:

v' = v(m/m')^(1/2)

Plugging in the values, we get:

v' = V($M/(M/2)$)^(1/2) = V/$\sqrt{2}$

Therefore, the new orbital speed is V/$\sqrt{2}$ when the mass of the planet is halved.

Question 2

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 3

It takes the planet Jupiter 12 years to orbit the sun once in a nearly circular orbit.Assuming that Jupiter's orbit is truly circular,what is the distance from Jupiter to the sun,given that the distance from the earth to the sun is 1.5 × 10¹¹ m? A)3.9 × 10¹¹ m B)5.2 × 10¹¹ m C)7.9 × 10¹¹ m D)9.7 × 10¹¹ m E)3.2 × 10¹² m

Accepted Answer

The answer of It takes the planet Jupiter 12 years...

Question 4

The planet Jupiter is 7.78 × 10¹¹ m from the sun.How long does it take for Jupiter to orbit once about the sun given that the distance from the earth to the sun is 1.50 × 10¹¹ m? A)2.3 years B)5.2 years C)6.0 years D)12 years E)27 years

Accepted Answer

The answer of The planet Jupiter is 7.78 × 10¹¹...

Question 5

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

The answer of The innermost satellite of Jupiter orbits the...

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 orbital diameter...

Question 7

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

To find the mass of the planet Zeeman-474, we can use the formula for the gravitational force acting as the centripetal force that keeps the satellite in orbit: $F = \frac{GmM}{r^2} = \frac{mv^2}{r}$, where $G$ is the gravitational constant, $m$ is the mass of the satellite, $M$ is the mass of the planet, $r$ is the distance from the center of the planet to the satellite, and $v$ is the orbital velocity of the satellite. However, since we are given the orbital period $T$ and not the velocity, we can use another form related to the orbital period: $T = 2\pi\sqrt{\frac{r^3}{GM}}$.Rearranging this formula to solve for $M$ gives us $M = \frac{4\pi^2r^3}{GT^2}$.Given:- $r =$ radius of the planet + altitude of the satellite = $6.38 \times 10^{6} m + 230 \times 10^{3} m = 6.61 \times 10^{6} m$- $T = 89$ minutes = $89 \times 60$ seconds (since we need $T$ in seconds)- $G = 6.67 \times 10^{-11} N \cdot m^2/kg^2$Plugging these values into the formula:$M = \frac{4\pi^2(6.61 \times 10^{6} m)^3}{(6.67 \times 10^{-11} N \cdot m^2/kg^2)(89 \times 60 s)^2}$This calculation yields a result close to $6.0 \times 10^{24} kg$, making option C the correct answer.

Question 8

The average distance from Earth to the sun is defined as one astronomical unit (AU).If an asteroid orbits the sun in 1/3 of a year in a circular orbit,the asteroid's distance from the sun is closest to
A)0.19 AU.
B)0.48 AU.
C)2.1 AU.
D)5.2 AU.
E)0.33 AU.

Accepted Answer

According to Kepler's Third Law, the square of the orbital period of a planet (or asteroid in this case) is directly proportional to the cube of the semi-major axis of its orbit. This law can be simplified to $T^2 \propto R^3$, where $T$ is the orbital period in years and $R$ is the semi-major axis in astronomical units (AU). If an asteroid orbits the sun in 1/3 of a year, we can set $T = 1/3$. To find $R$, we rearrange the formula to $R^3 = T^2$, giving us $R^3 = (1/3)^2$. Solving for $R$, we find $R \approx 0.48$ AU, making option B the closest answer.

Question 9

Two moons orbit a planet in circular orbits.Moon A has orbital radius R,and moon B has orbital radius 4R.Moon A takes 20 days to complete one orbit.How long does it take moon B to complete an orbit?
A)20 days
B)80 days
C)160 days
D)320 days
E)5 days

Accepted Answer

The orbital period of a moon (or any satellite) is determined by Kepler's third law, which states that the square of the orbital period (T) is directly proportional to the cube of the radius of its orbit (r), i.e., $T^2 \propto r^3$. If Moon A has an orbital radius R and takes 20 days to complete an orbit, and Moon B has an orbital radius of 4R, then the ratio of their periods squared is equal to the ratio of their radii cubed:$\frac{T_B^2}{T_A^2} = \frac{(4R)^3}{R^3} = 64$Given that $T_A = 20$ days, we can solve for $T_B$:$\frac{T_B^2}{20^2} = 64$$T_B^2 = 64 \times 400$$T_B^2 = 25600$$T_B = 160$ days.

Question 10

$$\begin{array} { | l | c | c | c | c | } \hline & \text { Mass } & \text { Radius } & \text { Orbital radius } & \text { Orbital period } \ \hline \text { Moon A } & 4.0 \times 10 ^ { 20 } \mathrm {~kg} & \text { unknown } & 2.0 \times 10 ^ { 8 } \mathrm {~m} & 4.0 \times 10 ^ { 6 } \mathrm {~s} \ \hline \text { Moon B } & 1.5 \times 10 ^ { 2 } 0 \mathrm {~kg} & 2.0 \times 10 ^ { 5 } \mathrm {~m} & 3.0 \times 10 ^ { 8 } \mathrm {~m} & \text { unknown } \ \hline \end{array}$$ Mithra is an unknown planet that has two 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 mass of Mithra can be calculated using the orbital period and radius of either moon, applying the formula for gravitational force and centripetal force. For Moon A, using the formula $T^2 = \frac{4\pi^2r^3}{GM}$, where $T$ is the orbital period, $r$ is the orbital radius, $G$ is the gravitational constant, and $M$ is the mass of Mithra, we can solve for $M$. Plugging in the values for Moon A, $T = 4.0 \times 10^6 \,s$ and $r = 2.0 \times 10^8 \,m$, we find that $M$ is approximately $3 \times 10^{23} \,kg$, which corresponds to choice D.

Question 11

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 $G$ is the gravitational constant $6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2$, $M$ is the mass of Ida $4.5 \times 10^{16} \, \text{kg}$, and $r$ is the distance between Dactyl and Ida $90 \, \text{km} = 90,000 \, \text{m}$. 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 12

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

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

Question 13

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 14

If NASA wants to put a satellite in a circular orbit around the sun so it will make 2.0 orbits per year,at what distance (in astronomical units,AU)from the sun should that satellite orbit? The earth's orbit is 1.0 AU from the sun.
A)7.6 AU
B)0.63 AU
C)0.50 AU
D)2.0 AU
E)0.71 AU

Accepted Answer

The period of an orbit (the time it takes to make one complete orbit) is related to the distance from the sun by Kepler's Third Law, which states that the square of the orbital period is directly proportional to the cube of the semi-major axis of its orbit. If the satellite makes 2.0 orbits per year, its period is 0.5 years (since 1 orbit/year is Earth's period). To maintain the same proportionality as Earth (1 AU, 1 year), the distance for a 0.5-year orbit would be the cube root of 0.5 squared (since T^2 ∝ R^3), which simplifies to the cube root of 0.25. This calculation gives a value close to 0.63 AU, making option B the correct choice.

Question 15

As you know,the earth has an orbital period of 1.0 year at an orbital radius of 1.0 AU (astronomical unit,the average distance between the earth and the sun).If a new "minor planet" were to be found in a circular orbit with radius 13 AU,what would be its orbital period?

Accepted Answer

The answer of As you know,the earth has an orbital...

Question 16

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 free-fall acceleration 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 because the planet's mass is 2.4 kg × 10^25 kg and the star's mass is 1.2 kg × 10^31 kg. This can be calculated using the formula for gravitational force, F = Gm1m2/r^2, where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

Question 17

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 18

If the earth were twice as far from the sun as it presently is,how long (in terms of the present year)would it take it to make one orbit around the sun?
A)2.0 years
B)2.8 years
C)4.0 years
D)5.7 years
E)16 years

Accepted Answer

According to Kepler's third law, the square of the orbital period is proportional to the cube of the semi-major axis of its orbit. If the distance is doubled, the period becomes √(2^3) = √8 ≈ 2.8 times the current period.

Question 19

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 square of the orbital period (T) of a satellite is directly proportional to the cube of its semimajor axis (a). We can express it as:
T^2 ∝ a^3

Since the altitude of the satellite is the same, we can assume that the semimajor axis remains the same. That is,
a1 = a2

Also, since the planet's mass remains the same, we can assume that its gravitational force on the satellite remains the same. That is,
F1 = F2

Now, the gravitational force (F) on the satellite is given by:
F = G*(M1*M2)/r^2

where G is the gravitational constant, M1 is the mass of the planet, M2 is the mass of the satellite, and r is the distance between their centers.

Since the altitude remains the same, we can assume that the distance between the centers remains the same. That is,
r1 = r2

Now, if the mass of the satellite is doubled (i.e., M2 = 2M), then the gravitational force becomes:
F2 = G*(M1*2M)/r^2
F2 = 2*(G*(M1*M)/r^2)
F2 = 2*F1

Since the force has doubled, the acceleration of the satellite (a=F2/M2) remains the same. Therefore, the only factor that changes the time period is the mass of the satellite. Using the above equation, we can write:
T1^2 ∝ M1
T2^2 ∝ 2M1

Dividing these equations, we get:
T2^2/T1^2 = 2
T2/T1 = √2

Therefore, the time period for the satellite with twice the mass is √2 times the time period for the original satellite. Thus, we get:
T2 = √2*T1

Since T2/T1 = √2, we can simplify it as:
T2 = T1*√2/√2
T2 = T1

Hence, 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. Therefore, the answer is C) T.

Quiz 5: Circular Motion

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

It takes the planet Jupiter 12 years to orbit the sun once in a nearly circular orbit.Assuming that Jupiter's orbit is truly circular,what is the distance from Jupiter to the sun,given that the distance from the earth to the sun is 1.5 × 10¹¹ m?

The planet Jupiter is 7.78 × 10¹¹ m from the sun.How long does it take for Jupiter to orbit once about the sun given that the distance from the earth to the sun is 1.50 × 10¹¹ m?

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)

The average distance from Earth to the sun is defined as one astronomical unit (AU) .If an asteroid orbits the sun in 1/3 of a year in a circular orbit,the asteroid's distance from the sun is closest to

Two moons orbit a planet in circular orbits.Moon A has orbital radius R,and moon B has orbital radius 4R.Moon A takes 20 days to complete one orbit.How long does it take moon B to complete an orbit?

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)

If NASA wants to put a satellite in a circular orbit around the sun so it will make 2.0 orbits per year,at what distance (in astronomical units,AU) from the sun should that satellite orbit? The earth's orbit is 1.0 AU from the sun.

As you know,the earth has an orbital period of 1.0 year at an orbital radius of 1.0 AU (astronomical unit,the average distance between the earth and the sun).If a new "minor planet" were to be found in a circular orbit with radius 13 AU,what would be its orbital period?

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)

If the earth were twice as far from the sun as it presently is,how long (in terms of the present year) would it take it to make one orbit around the sun?

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

Quiz 5: Circular Motion

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

It takes the planet Jupiter 12 years to orbit the sun once in a nearly circular orbit.Assuming that Jupiter's orbit is truly circular,what is the distance from Jupiter to the sun,given that the distance from the earth to the sun is 1.5 × 1011 m?

The planet Jupiter is 7.78 × 1011 m from the sun.How long does it take for Jupiter to orbit once about the sun given that the distance from the earth to the sun is 1.50 × 1011 m?

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)

The average distance from Earth to the sun is defined as one astronomical unit (AU) .If an asteroid orbits the sun in 1/3 of a year in a circular orbit,the asteroid's distance from the sun is closest to

Two moons orbit a planet in circular orbits.Moon A has orbital radius R,and moon B has orbital radius 4R.Moon A takes 20 days to complete one orbit.How long does it take moon B to complete an orbit?

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)

If NASA wants to put a satellite in a circular orbit around the sun so it will make 2.0 orbits per year,at what distance (in astronomical units,AU) from the sun should that satellite orbit? The earth's orbit is 1.0 AU from the sun.

As you know,the earth has an orbital period of 1.0 year at an orbital radius of 1.0 AU (astronomical unit,the average distance between the earth and the sun).If a new "minor planet" were to be found in a circular orbit with radius 13 AU,what would be its orbital period?

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)

If the earth were twice as far from the sun as it presently is,how long (in terms of the present year) would it take it to make one orbit around the sun?

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

It takes the planet Jupiter 12 years to orbit the sun once in a nearly circular orbit.Assuming that Jupiter's orbit is truly circular,what is the distance from Jupiter to the sun,given that the distance from the earth to the sun is 1.5 × 10¹¹ m?

The planet Jupiter is 7.78 × 10¹¹ m from the sun.How long does it take for Jupiter to orbit once about the sun given that the distance from the earth to the sun is 1.50 × 10¹¹ m?

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)

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 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)