Question 1

Two points on a merry-go-round are located at distances from the center r₁ and r₂, where r₁ < r₂.While the merry-go-round is in the process of speeding up to operational speed, which of the following equations involving magnitudes of angular speed, angular acceleration, and tangential speed for these points is incorrect? A)w₁ = w₂ B)a₁= a₂ C)v_t₁= v_t₂ D)All of the equations are correct.

Accepted Answer

The answer of Two points on a merry-go-round are located...

Question 2

For a point on a spinning disc in uniform circular motion, which of the following is not constant?
A)Its angular speed.
B)Its angular acceleration.
C)Its centripetal acceleration.
D)The magnitude of its total acceleration.

Accepted Answer

The answer of For a point on a spinning disc...

Question 3

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

The equation for gravitational acceleration is 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. Using the given values for Mars and Earth:

gMars = G * (0.107 * MEarth) / (0.530 * REarth)^2

gMars = (6.674 * 10^-11 N*m^2/kg^2) * (0.107 * 5.972 * 10^24 kg) / (0.530 * 6.371 * 10^6 m)^2

gMars = 3.73 m/s^2 (rounded to two decimal places)

Therefore, the best choice is B, as it is the closest value to the calculated value of 3.73 m/s^2.

Question 4

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 5

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 the period of revolution of a planet is proportional to the cube of its average distance from the Sun. Since the asteroid is at 4 times the distance of Earth from the Sun, its average distance is 4^3=64 times that of Earth's. Therefore, the period of revolution would be the square root of 64, which is 8 years.

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 law states that a planet or comet orbiting the Sun will sweep out equal areas in equal times, meaning that the object will move faster when it is closer to the Sun and slower when it is farther away. Because the orbit is elliptical, the object will speed up as it gets closer to the Sun and slow down as it moves away. The focus of the ellipse where the Sun is located is not the center of the ellipse, but rather one of the foci. The other focus does not have any physical object at its location.

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 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 8

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 9

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 10

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 a planet is proportional to the cube of the mean distance from the Sun. Therefore, doubling the mean distance from the Sun will result in a change in the orbital period by a factor of 2^(3/2) or approximately 2.828. Therefore, the correct answer is C, which represents this factor as a ratio of periods: 2^(3/2) or 2.828.

Question 11

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 12

Two objects are in circular orbits of different radii around the Sun.Which object has the highest orbital speed?
A)The one closest to the Sun.
B)The one farthest from the Sun.
C)Once in orbit around the Sun, all objects have the same orbital speed regardless of distance from the Sun.It is the greater radius and greater resulting circumference that causes the object farther from the Sun to take longer to complete an orbit.
D)This cannot be found without knowing the relative masses of the objects.

Accepted Answer

According to Kepler's laws, the orbital period of a planet is directly proportional to the cube of its semi-major axis (or distance from the Sun). Since the object closest to the Sun has a smaller radius (i.e. shorter distance from the Sun), it will have a shorter orbital period and therefore a higher orbital speed than the object farther away from the Sun.

Question 13

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

The period of orbital revolution for an Earth satellite is greater when the orbital radius is larger. This is because the period of revolution is directly proportional to the distance of the satellite from the center of the Earth (i.e. the orbital radius), according to Kepler's Third Law of Planetary Motion.

Question 14

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 formula a = v^2/r, where v is the velocity of the satellite and r is its distance from the center of the Earth. In an elliptical orbit, the distance from the Earth is constantly changing, but the velocity of the satellite is highest when it is closest to the Earth in its perigee. Therefore, the centripetal acceleration is greatest at this point.

Question 15

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 16

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

In order for a satellite to be geosynchronous, it must orbit the Earth once every 24 hours, which means it must be above a fixed point on the equator. Therefore, option B is correct. Option A is not correct because geosynchronous satellites do not orbit over the poles, instead they follow an elliptical path around the equator. Option C is not correct because a geosynchronous satellite can move slightly east or west in its orbit to maintain its position above a fixed point on the equator. Option D is not correct because emitting television signals is not a requirement for being geosynchronous.

Question 17

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...

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

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 force acting on the satellite can be found using the formula:
F = GMm/r^2
where G is the gravitational constant, 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.
Plugging in the values, we get:
F = (6.67 x 10^-11 Nm^2/kg^2)(5.97 x 10^24 kg)(1500 kg)/(42000 km)^2
Converting km to meters and simplifying, we get:
F = 3.33 x 10^2 N
Therefore, the correct answer is B) 333 N.

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

The aphelion can be found using the formula for the semi-major axis of an elliptical orbit, which is a = (r1 + r2)/2, where r1 and r2 are the distances from the focus (in this case, the sun) to the perihelion and aphelion, respectively. We can find r1 by subtracting the perihelion distance (1.5 AU) from the semi-major axis (which is also the average distance from the sun), so a = (1.5 AU + r2)/2. Multiplying both sides by 2 gives 2a = 1.5 AU + r2, and then subtracting 1.5 AU from both sides gives r2 (the aphelion distance) = 2a - 1.5 AU. We can find the value for a using Kepler's third law, which states that the square of the period of revolution is proportional to the cube of the semi-major axis: T^2/a^3 = constant. Rearranging this to solve for a gives a = (T^2)^(1/3). Substituting in the given values for T and solving gives a = 2.05 AU. Plugging this value into the previous equation for r2 gives r2 = (2.05 AU) - (1.5 AU) = 0.55 AU, or 6.5 AU. Therefore, the greatest distance from the sun, or aphelion, is 6.5 AU.

Quiz 7: Rotational Motion and the Law of Gravity

For a point on a spinning disc in uniform circular motion, which of the following is not constant?

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)

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?

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

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?

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?

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

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:

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

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?

Two objects are in circular orbits of different radii around the Sun.Which object has the highest orbital speed?

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

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

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

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

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.

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?

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) ?