Question 1

Which of the following best describes the property of the period of orbital revolution for an Earth satellite?
A) greater when the orbital radius is smaller
B) greater when the orbital radius is larger
C) independent of the orbital radius
D) determined mainly by the satellite's mass

Accepted Answer

According to Kepler's laws of planetary motion, the period of orbital revolution (the time it takes for a satellite to complete one orbit) is directly proportional to the radius of the orbit. This means that the period will be greater when the orbital radius is larger, making choice B the correct answer.

Question 2

In order for a satellite to be geosynchronous,its orbit must:
A) go over the North and South Poles.
B) be over the equator.
C) be over a single longitude.
D) emit television signals.

Accepted Answer

For a satellite to be geosynchronous, it must have an orbit around the Earth's equator at an altitude of about 35,786 kilometers. This allows the satellite to complete one orbit in the same amount of time that it takes for the Earth to rotate once on its axis, which is approximately 24 hours. This means that the satellite will appear to remain stationary over a specific location on the Earth's surface, making it ideal for communication and other applications.

Question 3

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 2 d?
A) 1 km/s
B) 2 km/s
C) 8 km/s
D) 16 km/s

Accepted Answer

According to Kepler's second law, the line connecting a planet to the sun sweeps out equal areas in equal intervals of time. This means that when the planet is closest to the sun, it moves faster and when it is farthest, it moves slower.

The conservation of angular momentum can also be applied to the asteroid's orbit:

angular momentum = moment of inertia x angular velocity

Since the moment of inertia of the asteroid is constant, when it is closer to the sun and moving faster, its angular velocity is higher. Therefore, when the asteroid is at its farthest distance, the angular velocity must be lower to maintain the same angular momentum.

Using the equation for angular velocity:

angular velocity = linear velocity / distance from the sun

we can calculate the new linear velocity when the asteroid is at a distance of 2d:

angular velocity at 2d = (4 km/s) / d

angular velocity at d = (v2) / (2d)

Setting the two equations equal and solving for v2, we get:

(4 km/s) / d = (v2) / (2d)

v2 = 8 km/s

Therefore, the asteroid's linear speed when it is at its greatest distance from the sun is 2 km/s.

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 1 500-kg satellite at this height?
A) 85 N
B) 333 N
C) 404 N
D) 457 N

Accepted Answer

The gravitational force acting on the satellite at this height can be calculated using the formula F = G(m1m2)/r^2, where G is the gravitational constant, m1 and m2 are the masses of the Earth and the satellite respectively, and r is the distance between their centers. Plugging in the values, we get F = (6.67x10^-11 Nm^2/kg^2)(1500 kg)(5.98x10^24 kg)/(42000 km)^2 = 332.5 N, which is closest to option B: 333 N.

Question 5

Two satellites are monitored as they orbit the Earth;satellite X is eight times as far from the Earth's center as is satellite Y.From Kepler's third law one may conclude that the period or revolution of X is what factor times that of Y?
A) 1/2
B) 2.0
C) 4.0
D) 22.6

Accepted Answer

The answer of Two satellites are monitored as they orbit...

Question 6

According to Kepler's second law,Halley's Comet circles the Sun in an elliptical path with the Sun at one focus of the ellipse.What is at the other focus of the ellipse?
A) nothing
B) the Earth
C) The comet itself passes through the other focus.
D) The tail of the comet stays at the other ellipse.

Accepted Answer

According to Kepler's second law, there is nothing at the other focus of the ellipse. The comet moves around the Sun in an elliptical path with the Sun at one focus and nothing at the other focus.

Question 7

Of the nine known planets in our solar system,the innermost is Mercury.When compared to the other planets in the system,Mercury has the:
A) greatest centripetal acceleration.
B) greatest period of revolution.
C) smallest angular velocity.
D) smallest tangential velocity.

Accepted Answer

Mercury, being the closest planet to the Sun, experiences the greatest centripetal acceleration due to its proximity and the gravitational pull from the Sun.

Question 8

Suppose the gravitational potential energy for a mass at an altitude of one Earth radius is PE₁ and at two Earth radii is PE₂.How do the absolute values of these potential energies compare? A) B) C) D) but is not equal to

Accepted Answer

The answer of Suppose the gravitational potential energy for a...

Question 9

The Earth's orbit is closest to the Sun in January and farthest from the Sun in July.When is the Earth moving the fastest in orbit?
A) Neither January nor July since the orbital speed of the Earth is a constant.
B) January
C) July
D) This occurs twice a year,in April and in October.

Accepted Answer

The Earth moves fastest in its orbit when it is closest to the Sun, due to the stronger gravitational pull at this point. This happens in January, making option B correct.

Question 10

At an altitude of 4 times the radius of the earth,the acceleration due to gravity is
A) g/2.
B) g/4.
C) g/16.
D) not given.

Accepted Answer

The answer of At an altitude of 4 times the...

Question 11

An asteroid is in orbit at 4 times the earth's distance from the Sun.What is its period of revolution?
A) one fourth year
B) 4 years
C) 8 years
D) 16 years

Accepted Answer

Kepler's third law states that the square of an object's period of revolution is proportional to the cube of its average distance from the Sun. Using this law, we can calculate the asteroid's period of revolution as follows:

(period of revolution)^2 = (average distance from Sun)^3
(period of revolution)^2 = (4)^3
(period of revolution)^2 = 64
period of revolution = sqrt(64)
period of revolution = 8

Therefore, the asteroid's period of revolution is 8 years, which is answer choice C.

Question 12

An artificial Earth satellite in an elliptical orbit has its greatest centripetal acceleration when it is at what location?
A) nearest the Earth
B) farthest from the Earth
C) between Earth and Moon
D) between Earth and Sun

Accepted Answer

Centripetal acceleration is given by the equation a=c²/r, where c is the speed of the satellite and r is the distance from the center of the Earth. Since the speed of the satellite is constant throughout its orbit, the centripetal acceleration is proportional to 1/r. At the point closest to the Earth, the satellite is at its smallest distance from the center of the Earth, and therefore experiences the greatest centripetal acceleration.

Question 13

If the mass of Mars is 0.107 times that of Earth,and its radius is 0.530 that of Earth,estimate the gravitational acceleration g at the surface of Mars.(g_earth = 9.80 m/s²) A) 2.20 m/s² B) 3.73 m/s² C) 4.20 m/s² D) 5.50 m/s²

Accepted Answer

We can use the formula for gravitational acceleration, g:

g = GM/r^2

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

We know that the mass of Mars is 0.107 times that of Earth, so:

M = 0.107 * M(earth)

Similarly, the radius of Mars is 0.530 that of Earth, so:

r = 0.530 * r(earth)

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

g = G * (0.107 * M(earth)) / (0.530 * r(earth))^2

g = (0.107/0.2809) * G * M(earth) / r(earth)^2

g = 0.381 * 9.80 m/s^2

g = 3.73 m/s^2

Therefore, the answer is B, 3.73 m/s^2.

Question 14

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

Accepted Answer

According to Kepler's second law, a satellite in elliptical orbit travels faster when it is closer to the Earth. Therefore, the artificial Earth satellite will have its greatest speed at the point nearest to the Earth.

Question 15

A satellite is in a circular orbit about the Earth at a distance of one Earth radius 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) 4 200 m/s C) 5 600 m/s D) 16 800 m/s

Accepted Answer

The speed of the satellite in a circular orbit is given by the formula:
v = √(GM/r)
Where G is the gravitational constant, M is the mass of the Earth, and r is the distance between the satellite and the center of the Earth.
Substituting the given values, we get:
v = √((6.67 × 10−11 N⋅m2/kg2) × (5.97 × 10^24 kg) / (6.4 × 10^6 m))
v = 7.91 × 10^3 m/s
However, this is the speed of the satellite with respect to the center of the Earth. To find the speed of the satellite with respect to the surface of the Earth, we need to subtract the speed of the surface of the Earth at the satellite's altitude, which is approximately 7.91 × 10^3 m/s. Therefore, the speed of the satellite with respect to the surface of the Earth is approximately:
v = 7.91 × 10^3 m/s - 7.91 × 10^3 m/s = 0
This means that the satellite is moving at the same speed as the surface of the Earth at that altitude, which is zero. Therefore, the answer is (C) 5 600 m/s.

Question 16

A careful photographic survey of Jupiter's moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to heights of 70 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) 260 m/s B) 530 m/s C) 790 m/s D) 970 m/s

Accepted Answer

The answer of A careful photographic survey of Jupiter's moon...

Question 17

Doubling the mean distance from the Sun results in changing the orbital period of revolution by what factor? A) 2^1/2 B) 2 C) 2^3/2 D) 2²

Accepted Answer

According to Kepler's Third Law, the square of the orbital period of revolution of a planet around the Sun is directly proportional to the cube of the mean distance from the Sun. If the mean distance is doubled, then the orbital period of revolution will be multiplied by the square root of (2^3) or 2^(3/2), which is approximately 2.83 (rounded to two decimal places). Therefore, the correct answer is C: 2^(3/2), or 2 times the square root of 2.

Question 18

The escape speed from the surface of the Earth is 11.2 km/s.Estimate the escape speed for a spacecraft from the surface of the Moon.The Moon has a mass 1/81 that of Earth and a radius 0.25 that of Earth.
A) 2.5 km/s
B) 4.0 km/s
C) 5.6 km/s
D) 8.7 km/s

Accepted Answer

The escape velocity $v$ from a celestial body is given by the formula $v = \sqrt{\frac{2GM}{r}}$, where $G$ is the gravitational constant, $M$ is the mass of the celestial body, and $r$ is its radius. Given that the Moon's mass is $1/81$ that of Earth and its radius is $0.25$ times that of Earth, the escape velocity from the Moon can be calculated relative to Earth's escape velocity ($11.2$ km/s) as follows: $v_{\text{Moon}} = 11.2 \times \sqrt{\frac{1/81}{0.25^2}} = 11.2 \times \sqrt{\frac{1}{81}} \times \sqrt{16} = 11.2 \times \frac{1}{9} \times 4 = 2.5$ km/s.

Question 19

For any object orbiting the Sun,Kepler's Law may be written T² = kr³.If T is measured in years and r in units of the Earth's distance from the Sun,then k = 1.What,therefore,is the time (in years)for Mars to orbit the Sun if its mean radius from the Sun is 1.5 times the Earth's distance from the Sun? A) 1.8 years B) 2.8 years C) 3.4 years D) 4.2 years

Accepted Answer

The answer of For any object orbiting the Sun,Kepler's Law...

Question 20

An asteroid has a perihelion (the orbit's closest approach to the sun)of 1.5 AU and a period of revolution of 8.0 y.What is its greatest distance from the sun (its aphelion)?
A) 8.0 AU
B) 6.5 AU
C) 4.0 AU
D) 2.5 AU

Accepted Answer

We can use Kepler's laws of planetary motion to solve this problem. The third law states that the ratio of the square of a planet's period of revolution to the cube of its average distance from the sun is a constant for all planets in the solar system. Mathematically, it can be expressed as:

(T1/T2)^2 = (a1/a2)^3

where T is the period of revolution and a is the average distance from the sun. We can rearrange this equation to solve for a2, the aphelion distance:

a2 = a1(T2/T1)^(2/3)

In this case, a1 = 1.5 AU (the perihelion distance), T1 = 8.0 years, and we want to find a2. The period T2 is the same as T1, since the asteroid's orbit is fixed, so we can plug in the values:

a2 = 1.5 AU (8.0 years / 8.0 years)^(2/3) = 6.5 AU

Therefore, the asteroid's greatest distance from the sun (its aphelion) is 6.5 AU.

Quiz 7: Rotational Motion and the Law of Gravity

Which of the following best describes the property of the period of orbital revolution for an Earth satellite?

In order for a satellite to be geosynchronous,its orbit must:

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 2 d?

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 1 500-kg satellite at this height?

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

According to Kepler's second law,Halley's Comet circles the Sun in an elliptical path with the Sun at one focus of the ellipse.What is at the other focus of the ellipse?

Of the nine known planets in our solar system,the innermost is Mercury.When compared to the other planets in the system,Mercury has the:

Suppose the gravitational potential energy for a mass at an altitude of one Earth radius is PE₁ and at two Earth radii is PE₂.How do the absolute values of these potential energies compare?

The Earth's orbit is closest to the Sun in January and farthest from the Sun in July.When is the Earth moving the fastest in orbit?

At an altitude of 4 times the radius of the earth,the acceleration due to gravity is

An asteroid is in orbit at 4 times the earth's distance from the Sun.What is its period of revolution?

An artificial Earth satellite in an elliptical orbit has its greatest centripetal acceleration when it is at what location?

If the mass of Mars is 0.107 times that of Earth,and its radius is 0.530 that of Earth,estimate the gravitational acceleration g at the surface of Mars.(g_earth = 9.80 m/s²)

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

A satellite is in a circular orbit about the Earth at a distance of one Earth radius 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 careful photographic survey of Jupiter's moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to heights of 70 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.

Doubling the mean distance from the Sun results in changing the orbital period of revolution by what factor?

The escape speed from the surface of the Earth is 11.2 km/s.Estimate the escape speed for a spacecraft from the surface of the Moon.The Moon has a mass 1/81 that of Earth and a radius 0.25 that of Earth.

An asteroid has a perihelion (the orbit's closest approach to the sun) of 1.5 AU and a period of revolution of 8.0 y.What is its greatest distance from the sun (its aphelion) ?