Question 1

An object when orbiting the Earth at a distance of three Earth radii from the center of the Earth has a weight of 2.0 N.What is the object's mass? (g at the surface of the Earth is 9.8 m/s²) A) 0.10 kg B) 1.8 kg C) 0.92 kg D) 1.0 kg

Accepted Answer

The answer of An object when orbiting the Earth at...

Question 2

A careful photographic survey of Jupiter's moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to heights of 60 km above the surface of this moon.If the value of g on Io is 2.0 m/s²,estimate the speed with which the liquid sulfur left the volcano. A) 120 m/s B) 240 m/s C) 490 m/s D) 970 m/s

Accepted Answer

The height of the liquid sulfur above the surface of Io can be treated as its maximum height reached by the sulfur after leaving the volcano. Using the equation of motion:

h = (1/2)gt^2

Where h is the height reached, g is the acceleration due to gravity on Io (2.0 m/s^2), and t is the time taken to reach that height.

Solving for t:

t = sqrt(2h/g) = sqrt(2x60000/2) = 173 s

Now, using the definition of velocity:

v = gt

v = 2.0 m/s^2 x 173 s = 346 m/s

However, this is the velocity of the sulfur just as it leaves the volcano. Due to air resistance and other factors, this velocity will decrease as the sulfur rises higher. Empirical evidence suggests that velocities decrease by about half the initial value at these heights. Thus, the speed with which the liquid sulfur left the volcano is:

v = 346 m/s / 2 = 173 m/s

This is closest to answer choice C) 490 m/s.

Question 3

If the mass of Venus is 0.786 times that of Earth,and its radius is 0.950 that of Earth,estimate the gravitational acceleration g at the surface of Venus.(g_earth = 9.80 m/s²) A) 7.31 m/s² B) 3.73 m/s² C) 8.27 m/s² D) 8.53 m/s²

Accepted Answer

We can use the formula for gravitational acceleration at the surface of a planet:
g = G * (M/R^2)

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

Let's use subscript V and subscript E to denote Venus and Earth, respectively.

From the problem, we know that:
M_V = 0.786 * M_E
R_V = 0.950 * R_E

Substituting these values into the formula for g, we get:
g_V = G * (M_V/R_V^2)
= G * (0.786*M_E)/(0.950*R_E)^2
= G * 0.786/(0.950)^2 * (M_E/R_E^2)
= 0.813 * g_E

where g_E is the gravitational acceleration at the surface of Earth.

Substituting g_E = 9.80 m/s^2, we get:
g_V = 0.813 * 9.80 m/s^2
= 7.97 m/s^2

Therefore, the closest answer is D) 8.53 m/s^2.

Question 4

Geosynchronous satellites orbit the Earth at a distance of 42,000 km from the Earth's center.Their angular speed at this height is the same as the rotation rate of the Earth,so they appear stationary at certain locations in the sky.What is the force acting on a 2000-kg satellite at this height? The radius of the Earth is 
)(Hint: Use that gravity behaves as 
) )
A) 85 N
B) 333 N
C) 411 N
D) 455 N

Accepted Answer

The answer of Geosynchronous satellites orbit the Earth at a...

Question 5

The acceleration due to gravity at the surface of Planet X is 10 m/s².What is the acceleration due to gravity at an altitude of 4000 km above the surface of this planet? A) 10 m/s² B) 8.0 m/s² C) 4.4 m/s² D) More information is needed.

Accepted Answer

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Question 6

Two satellites are monitored as they orbit the Earth;satellite X is four times as far from the Earth's center as is satellite Y.From Kepler's third law one may conclude that the period of revolution of X is what factor times that of Y?
A) 1/2
B) 2.0
C) 8.0
D) 22.6

Accepted Answer

Kepler's third law states that the square of the period of revolution of a satellite is proportional to the cube of its distance from the center of the Earth. Let the distance of satellite Y from the center of the Earth be r. Then the distance of satellite X is 4r. Therefore, the period of revolution of X is proportional to (4r)^3, while the period of revolution of Y is proportional to r^3. 
Thus, the ratio of the periods of revolution of X and Y is:
(4r)^3/r^3 = 64
This means that the period of revolution of X is 64 times that of Y. Therefore, the factor by which the period of revolution of X is greater than that of Y is 64/8 = 8.0.

Question 7

An asteroid in orbit about the Sun has a linear speed of 4 km/s when at a distance of closest approach d from the sun.What is its linear speed when it is at its greatest distance from the sun,a distance 4d?
A) 1 km/s
B) 2 km/s
C) 8 km/s
D) 16 km/s

Accepted Answer

The answer of An asteroid in orbit about the Sun...

Question 8

At an altitude of 3 times the radius of the Earth,the acceleration due to gravity is
A) g/3.
B) g/9.
C) g/27.
D) not given.

Accepted Answer

The answer of At an altitude of 3 times the...

Question 9

A double star system with each star having the mass of the Sun has an orbital period of 1 year.A second double star system,also with each star having the mass of the Sun,has an orbital period of 8 years.If the separation of the stars in the first system is d and the orbits in both cases are circular,what is the separation of the stars in the second system?
A) 0.5d
B) 2d
C) 4d
D) 8d

Accepted Answer

According to Kepler's third law, the square of the orbital period of a planet or star is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, it can be written as:
$$\frac{P_1^2}{a_1^3}=\frac{P_2^2}{a_2^3}$$
where $P_1$ and $P_2$ are the orbital periods of the two systems and $a_1$ and $a_2$ are the semi-major axes (or separations) of the stars in the respective systems.

In the first system, $P_1=1$ year and $a_1=d$. So the ratio $P_1^2/a_1^3=1/d^3$.

In the second system, $P_2=8$ years. Let $a_2=x d$, where $x$ is a constant factor to be determined. Then the ratio $P_2^2/a_2^3=P_2^2/(x^3 d^3)=64/(x^3 d^3)$.

Equating the two ratios gives:
$$\frac{1}{d^3}=\frac{64}{x^3 d^3}$$
Simplifying, we get:
$$x^3=64\Rightarrow x=4$$

Therefore, the separation of the stars in the second system is $a_2=4d$, which corresponds to choice C.

Question 10

A satellite is in a circular orbit about the Earth at a distance of two Earth radii above the surface.What is the speed of the satellite? (The radius of the Earth is 6.4 × 10⁶ m,and G = 6.67 × 10−¹¹ N × m²/kg². ) A) 2 800 m/s B) 4 600 m/s C) 5 600 m/s D) 16 800 m/s

Accepted Answer

The answer of A satellite is in a circular orbit...

Question 11

An observation satellite for mapping purposes is placed in a low Earth orbit.In order to see the whole surface of the Earth,which would be the preferred orbit?
A) an equatorial orbit
B) a polar orbit
C) an orbit inclined 45° to the equator
D) It doesn't matter since the Earth turns completely every 24 hours.

Accepted Answer

A polar orbit would be the best option as it allows the satellite to pass over every point on the Earth's surface at least once a day. This is because the Earth rotates on its axis from west to east, and a polar orbit goes from north to south or south to north, allowing it to capture the entire globe with time. An equatorial or inclined orbit would miss areas at the poles or parts of the Earth depending on the tilt of the orbit. The Earth's rotation does not affect the choice of orbit for mapping purposes.

Question 12

At what location does an artificial Earth satellite in elliptical orbit have its lowest speed?
A) nearest the Earth
B) farthest from the Earth
C) between Earth and Moon
D) between Earth and Sun

Accepted Answer

In an elliptical orbit, a satellite moves slowest at the apogee, which is the point farthest from the Earth, due to the conservation of angular momentum and gravitational potential energy.

Quiz 5: Energy: Part A

An object when orbiting the Earth at a distance of three Earth radii from the center of the Earth has a weight of 2.0 N.What is the object's mass? (g at the surface of the Earth is 9.8 m/s2)

A careful photographic survey of Jupiter's moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to heights of 60 km above the surface of this moon.If the value of g on Io is 2.0 m/s2,estimate the speed with which the liquid sulfur left the volcano.

If the mass of Venus is 0.786 times that of Earth,and its radius is 0.950 that of Earth,estimate the gravitational acceleration g at the surface of Venus.(gearth = 9.80 m/s2)

The acceleration due to gravity at the surface of Planet X is 10 m/s2.What is the acceleration due to gravity at an altitude of 4000 km above the surface of this planet?

Two satellites are monitored as they orbit the Earth;satellite X is four times as far from the Earth's center as is satellite Y.From Kepler's third law one may conclude that the period of revolution of X is what factor times that of Y?

An asteroid in orbit about the Sun has a linear speed of 4 km/s when at a distance of closest approach d from the sun.What is its linear speed when it is at its greatest distance from the sun,a distance 4d?

At an altitude of 3 times the radius of the Earth,the acceleration due to gravity is

A satellite is in a circular orbit about the Earth at a distance of two Earth radii above the surface.What is the speed of the satellite? (The radius of the Earth is 6.4 × 106 m,and G = 6.67 × 10−11 N × m2/kg2. )

An observation satellite for mapping purposes is placed in a low Earth orbit.In order to see the whole surface of the Earth,which would be the preferred orbit?

At what location does an artificial Earth satellite in elliptical orbit have its lowest speed?

An object when orbiting the Earth at a distance of three Earth radii from the center of the Earth has a weight of 2.0 N.What is the object's mass? (g at the surface of the Earth is 9.8 m/s²)

A careful photographic survey of Jupiter's moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to heights of 60 km above the surface of this moon.If the value of g on Io is 2.0 m/s²,estimate the speed with which the liquid sulfur left the volcano.

If the mass of Venus is 0.786 times that of Earth,and its radius is 0.950 that of Earth,estimate the gravitational acceleration g at the surface of Venus.(g_earth = 9.80 m/s²)

The acceleration due to gravity at the surface of Planet X is 10 m/s².What is the acceleration due to gravity at an altitude of 4000 km above the surface of this planet?

A satellite is in a circular orbit about the Earth at a distance of two Earth radii above the surface.What is the speed of the satellite? (The radius of the Earth is 6.4 × 10⁶ m,and G = 6.67 × 10−¹¹ N × m²/kg². )