Question 1

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

We can use the formula for maximum height reached by a projectile:
h = (v²sin²θ)/(2g)
where v is the initial velocity of the projectile, θ is the angle of projection (which we can assume to be 90° in this case), and g is the acceleration due to gravity.

Rearranging the formula, we get:
v = sqrt((2gh)/(sin²θ))

Substituting the given values, we get:
v = sqrt((2 x 2.0 x 70 x 10³)/(1))
v ≈ 529.15 m/s ≈ 530 m/s

So, the speed with which the liquid sulfur left the volcano is approximately 530 m/s, which is closest to option B.

Question 2

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, the other focus of an elliptical orbit is empty. There is nothing present at that point.

Question 3

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

The answer of Of the nine known planets in our...

Question 4

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 Third Law, the period of a satellite's orbit is directly proportional to the radius of the orbit raised to the 3/2 power. Therefore, the period of revolution will be greater when the radius of the orbit is larger.

Question 5

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

Geosynchronous means that the satellite orbits the Earth at the same rate that the Earth rotates on its axis, which is once per day. This requires the satellite to be positioned over the equator.

Question 6

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 7

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

According to Kepler's Second Law, a satellite moves fastest (and therefore has the greatest centripetal acceleration) when it is closest to the body it is orbiting. Therefore, the artificial Earth satellite will have its greatest centripetal acceleration when it is nearest the Earth.

Question 8

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 x 10⁶ m,and G = 6.67 x 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 a satellite in a circular orbit is given by the formula:

v = sqrt(GM/R)

Where:
G = gravitational constant = 6.67 x 10^-11 N m^2/kg^2
M = mass of the Earth = 5.97 x 10^24 kg
R = radius of the orbit = (radius of the Earth + altitude of the orbit) = (6.4 x 10^6 m + 6.4 x 10^6 m) = 1.28 x 10^7 m

Substituting the values, we get:

v = sqrt(6.67 x 10^-11 N m^2/kg^2 x 5.97 x 10^24 kg / 1.28 x 10^7 m)
v = sqrt(2.98 x 10^14 m^3/s^2)
v = 5.6 x 10^3 m/s

Therefore, the best choice is C) 5 600 m/s.

Question 9

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

We can use the formula for gravitational force:

F = G(m1m2)/r^2

where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

At the altitude of 42,000 km, the satellite is still within the Earth's gravitational field, but the force of gravity is weaker than at the Earth's surface. We can approximate the distance between the satellite and the Earth's center as 42,000 km, and assume that the mass of the Earth is concentrated at the center.

Then, the force acting on the satellite would be:

F = G(Mm)/r^2

where M is the mass of the Earth, m is the mass of the satellite, and r is the distance between the center of the Earth and the satellite.

We can plug in the given values:

F = (6.67 x 10^-11 Nm^2/kg^2)(5.97 x 10^24 kg)(1,500 kg)/(42,000,000 m)^2

F = 333 N

Therefore, the best answer is B.

Question 10

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

First, 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.

We are given that the mass of Mars (M) is 0.107 times that of Earth, and the radius of Mars (R) is 0.530 that of Earth.

So,

g = G((0.107)(Mearth)/((0.53)(Rearth))^2)

We can simplify this by noting that (0.107)/(0.53^2) is approximately 0.37. Then, using the value G = 6.67 x 10^-11 N(m/kg)^2 and the fact that Mearth = 5.97 x 10^24 kg and Rearth = 6.37 x 10^6 m, we get:

g = (6.67 x 10^-11)(0.37)(5.97 x 10^24)/(6.37 x 10^6)^2

g ≈ 3.73 m/s^2

Therefore, the best choice is B.

Question 11

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 12

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 orbital period of a planet squared is proportional to the cube of its mean distance from the Sun. Doubling the mean distance from the Sun results in a cube that is 2^3 = 8 times larger. Taking the square root of 8 gives a factor of 2.83. Therefore, the orbital period of revolution is multiplied by 2.83, which is represented by choice C: 2^3/2.

Question 13

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 14

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 15

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 16

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 17

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 18

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

Using Kepler's Third Law, we can calculate the asteroid's semi-major axis: a^3 = P^2, where a is the semi-major axis in AU and P is the period in years. Thus, a^3 = 8^2, or a = 2 AU. The aphelion distance is equal to the semi-major axis plus the distance at perihelion: a + q = 2 AU + 1.5 AU = 3.5 AU. Therefore, the greatest distance from the sun is the aphelion distance, which is 3.5 AU multiplied by 2 or 7.0 AU (since the asteroid's orbit is elliptical, we know the distance varies) divided by 2 equals 6.5 AU.

Question 19

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

An artificial Earth satellite in elliptical orbit has its greatest speed when it is nearest the Earth, due to the stronger gravitational pull at this point, which is described by Kepler's second law (the law of areas).

Question 20

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...

Quiz 7: Rotational Motion and the Law of Gravity

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.

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:

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:

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?

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

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 x 10⁶ m,and G = 6.67 x 10^-11 N*m²/kg². )

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?

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

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

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?

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

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?

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?

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) ?

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