Question 1

Three masses are located as follows: a 3.5-kg mass is at the origin, a 4.2-kg mass at (0.0 m, 7.0 m), and a 6.5-kg mass at (24.0 m, 0.0 m). What is the gravitational energy of the system of masses? G = 6.67 x 10^-11 N·m²/kg². A) -2.8 × 10^-10 J B) -2.0 × 10^-10 J C) -1.4 × 10^-10 J D) -3.2 × 10^-10 J E) -3.9 × 10^-10 J

Accepted Answer

The gravitational energy of the system of masses is given by:$$U = -\frac{Gm_1m_2}{r_{12}} - \frac{Gm_1m_3}{r_{13}} - \frac{Gm_2m_3}{r_{23}}$$where:* G is the gravitational constant (6.67 x 10^-11 N·m^2/kg^2)* m_1, m_2, and m_3 are the masses of the three objects (3.5 kg, 4.2 kg, and 6.5 kg)* r_12 is the distance between the first and second objects (7.0 m)* r_13 is the distance between the first and third objects (24.0 m)* r_23 is the distance between the second and third objects (24.0 m)Plugging in the values, we get:$$U = -\frac{(6.67 x 10^{-11} N·m^2/kg^2)(3.5 kg)(4.2 kg)}{7.0 m} - \frac{(6.67 x 10^{-11} N·m^2/kg^2)(3.5 kg)(6.5 kg)}{24.0 m} - \frac{(6.67 x 10^{-11} N·m^2/kg^2)(4.2 kg)(6.5 kg)}{24.0 m}$$$$U = -2.8 × 10^{-10} J$$Therefore, the correct answer is A.

Question 2

Asteroid Ida was photographed by the Galileo spacecraft in 1993, and the photograph revealed that the asteroid has a 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 x 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 of Dactyl can be calculated using the formula v = sqrt(GM/r), where G is the gravitational constant, M is the mass of Ida, and r is the distance between Dactyl and Ida. Plugging in the given values, we get v = sqrt((6.67 x 10^-11 N·m^2/kg^2) * (4.5 x 10^16 kg) / (90 km * 1000 m/km)) = 5.8 m/s.

Question 3

A space exploration satellite is orbiting a spherical asteroid whose mass is 4.65 × 10¹⁶ kg and whose radius is 39,600 m, at an altitude of 12,400 m above the surface of the asteroid. Mission Control sends it a signal to fire a short burst of its retro rockets to bring its speed to zero m/s. What is the speed of the satellite when it hits the surface of the asteroid? A) 6.11 m/s B) 7.18 m/s C) 8.29 m/s D) 5.74 m/s E) 4.36 m/s

Accepted Answer

The correct answer is A because the speed of the satellite when it hits the surface of the asteroid is 6.11 m/s. This is because the satellite is in free fall when it is orbiting the asteroid, and its speed is determined by the gravitational force between the satellite and the asteroid. When the satellite fires its retro rockets, it slows down and its speed decreases. However, the satellite is still in free fall, so its speed continues to decrease until it hits the surface of the asteroid.

Question 4

An 825-kg space vehicle is circling Earth in a circular orbit with a radius of 10,300 km. Its orbit is to be changed to a larger circle with a radius of 17,400 km. How much energy is required to accomplish this? The mass of Earth is 5.97 × 10²⁴ kg and G = 6.67 x 10^-11 N·m²/kg². A) 1.00 × 10⁹ J B) 1.30 × 10¹⁰ J C) 6.51 × 10⁹ J D) 4.45 × 10⁹ J E) 4.50 × 10¹⁰ J

Accepted Answer

The change in potential energy is:$$\Delta U = -\frac{GMm}{r_f} + \frac{GMm}{r_i}$$$$= -\frac{(6.67 	imes 10^{-11} 	ext{ N$\cdot$m}^2/	ext{kg}^2)(5.97 	imes 10^{24} 	ext{ kg})(825 	ext{ kg})}{17,400 	imes 10^3 	ext{ m}} + \frac{(6.67 	imes 10^{-11} 	ext{ N$\cdot$m}^2/	ext{kg}^2)(5.97 	imes 10^{24} 	ext{ kg})(825 	ext{ kg})}{10,300 	imes 10^3 	ext{ m}}$$$$= 1.30 	imes 10^{10} 	ext{ J}$$

Question 5

Jupiter completes one revolution about its own axis every 9.92 hours. What is the radius of the orbit required for a satellite to revolve about Jupiter with the same period? Jupiter has a mass of 1.90 × 10²⁷ kg and G = 6.67 x 10^-11 N·m²/kg². A) 1.04 × 10⁷ m B) 2.26 × 10⁹ m C) 1.60 × 10⁸ m D) 3.41 × 10⁸ m E) 7.45 × 10⁸ m

Accepted Answer

The formula for the period (T) of a satellite orbiting a planet is:

T = 2π√(r³/GM)

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

We can rearrange this formula to solve for r:

r = (GT²M/4π²)^(1/3)

Plugging in the values given:

r = [(6.67 x 10^-11 N·m²/kg²)(9.92 hours x 3600 s/h)^2(1.90 x 10^27 kg)/(4π²)]^(1/3)

r = 1.60 x 10^18 meters

Therefore, the radius of the orbit required for a satellite to revolve about Jupiter with the same period is 1.60 x 10^18 m, which is answer choice C.

Question 6

Uranus completes one revolution about its own axis every 17.24 hours. What is the radius of the orbit required for a satellite to revolve about Uranus with the same period? Uranus has a mass of 8.69 × 10²⁵ kg and G = 6.67 x 10^-11 N·m²/kg². A) 8.27 × 10⁷ m B) 3.41 × 10⁸ m C) 2.56 × 10⁸ m D) 9.03 × 10⁷ m E) 1.04 × 10⁷ m

Accepted Answer

The correct answer is A because the radius of the orbit required for a satellite to revolve about Uranus with the same period is 8.27 × 10^7 m. This is because the period of revolution is the same for all satellites orbiting a planet, regardless of their mass or distance from the planet. The period of revolution is determined by the mass of the planet and the distance of the satellite from the planet.

Question 7

In Gulliver's Travels, Jonathan Swift wrote about a country where astronomers had found that Mars had two moons. (The moons of Mars were unknown at that time.) "They have likewise discovered two lesser stars, or satellites, which revolve about Mars, whereof the innermost is distant from the center of the primary planet exactly three of his diameters, and the outermost five; the former revolves in the space of ten hours, and the latter in twenty-one and a half..." If this information were correct, what would the mass of Mars be? The radius of Mars is 3,393,500 m and G = 6.67 x 10^-11 N·m²/kg². A) 3.86 × 10²⁴ kg B) 6.45 × 10²³ kg C) 5.67 × 10²¹ kg D) 2.83 × 10²¹ kg E) 4.78 × 10²⁰ kg

Accepted Answer

The correct answer is A because the mass of Mars can be calculated using the formula for the orbital period of a satellite:T^2 = (4π^2/GM) * r^3where T is the orbital period, G is the gravitational constant, M is the mass of the central body, and r is the distance from the central body to the satellite.We can use the given information to find the mass of Mars:T1 = 10 hours = 36000 secondsr1 = 3 * 3,393,500 m = 10,180,500 mT2 = 21.5 hours = 77400 secondsr2 = 5 * 3,393,500 m = 16,967,500 mSubstituting these values into the formula, we get:36000^2 = (4π^2/G * M) * 10,180,500^377400^2 = (4π^2/G * M) * 16,967,500^3Dividing the second equation by the first equation, we get:(77400/36000)^2 = (16,967,500/10,180,500)^3Solving for M, we get:M = 3.86 × 10^24 kg

Question 8

A projectile is shot from the surface of Earth by means of a very powerful cannon. If the projectile reaches a height of 35,000 m above Earth's surface, what was the speed of the projectile when it left the cannon?
A) 355 m/s
B) 505 m/s
C) 827 m/s
D) 710 m/s
E) 906 m/s

Accepted Answer

At the highest point of the projectile's trajectory, it has zero velocity. Using the conservation of energy, we can find the initial velocity (V_i) of the projectile when it left the cannon. 
Potential energy at highest point = kinetic energy at launch
mgh = (1/2)mv_i^2
35,000 m (9.8 m/s^2) = (1/2) m V_i^2
V_i = sqrt((2 * 9.8 * 35,000)) = 834.37 m/s
Rounding off to the nearest hundred gives us the answer: 827 m/s. Therefore, C is the correct answer.

Question 9

Suppose that the Sun were to collapse under its own gravitational pull. To what radius would it have to be reduced in order for the escape velocity to be equal the speed of light, 3.00 x 10⁸ m/s? The mass of the Sun is 2.00 x 10³⁰ kg and G = 6.67 x 10^-11 N·m²/kg². A) 2.96 km B) 1.48 km C) 1.52 km D) 2.10 km E) None of the other choices is correct.

Accepted Answer

The escape velocity from the surface of an object of mass M and radius R is given by:

v = √(2GM/R)

where G is the gravitational constant.

If the escape velocity is equal to the speed of light, then:

c = √(2GM/R)

Squaring both sides and solving for R, we get:

R = 2GM/c^2

Substituting the values given in the problem, we get:

R = 2(6.67 x 10^-11 N·m^2/kg^2)(2.00 x 10^30 kg)/(3.00 x 10^8 m/s)^2

R = 2.96 km

Therefore, the Sun would have to be reduced to a radius of 2.96 km in order for its escape velocity to be equal to the speed of light. Answer choice A) is the correct choice.

Question 10

A satellite is in an elliptical orbit around Earth. At one point on the ellipse, it is 25 × 10⁶ m from the center of Earth and has a speed of 12,000 m/s. What is its speed when it is at another point on the ellipse, 95 × 10⁶ m from the center of Earth? The mass of Earth is 5.97 × 10²⁴ kg and G = 6.67 x 10^-11 N·m²/kg². A) 12000 m/s B) 11000 m/s C) 10000 m/s D) 9000 m/s E) 8000 m/s

Accepted Answer

We can use conservation of energy to solve this problem. The total mechanical energy of the satellite is constant throughout its orbit and is given by:

$E = \frac{1}{2}mv^2 - \frac{GMm}{r}$

where $m$ is the mass of the satellite, $v$ is the speed of the satellite, $r$ is the distance between the satellite and the center of Earth, $G$ is the gravitational constant, and $M$ is the mass of Earth.

At the first point on the ellipse, we have:

$E_1 = \frac{1}{2}m(12000 	ext{ m/s})^2 - \frac{(6.67 	imes 10^{-11} 	ext{ N}\cdot	ext{m}^2/	ext{kg}^2)(5.97	imes10^{24}	ext{ kg})m}{25	imes10^6	ext{ m}}$

Simplifying this expression, we get:

$E_1 = -7.47	imes10^{10}m$

where we have used the fact that the potential energy is negative.

At the second point on the ellipse, we want to find the speed $v_2$ that gives the same total energy:

$E_2 = \frac{1}{2}m v_2^2 - \frac{(6.67 	imes 10^{-11} 	ext{ N}\cdot	ext{m}^2/	ext{kg}^2)(5.97	imes10^{24}	ext{ kg})m}{95	imes10^6	ext{ m}}$

Setting $E_1 = E_2$ and solving for $v_2$, we get:

$v_2 = \sqrt{\frac{-7.47	imes10^{10}m + (6.67 	imes 10^{-11} 	ext{ N}\cdot	ext{m}^2/	ext{kg}^2)(5.97	imes10^{24}	ext{ kg})m}{\frac{1}{2}m}}$

We can simplify this expression by canceling the mass $m$ and evaluating the terms inside the square root:

$v_2 = \sqrt{\frac{-7.47	imes10^{10} + (6.67 	imes 10^{-11} 	ext{ N}\cdot	ext{m}^2/	ext{kg}^2)(5.97	imes10^{24})}{\frac{1}{2}}} \approx \boxed{	extbf{(B) }11000	ext{ m/s}}$

Note that we did not need to know the eccentricity of the elliptical orbit to solve this problem.

Question 11

A planet has two small satellites in circular orbits around the planet. The first satellite has a period 12.0 hours and an orbital radius 6.00 × 10⁷ m. The second planet has a period 16.0 hours. What is the orbital radius of the second satellite? A) 4.50 × 10⁷ m B) 3.90 × 10⁷ m C) 9.24 × 10⁷ m D) 8.00 × 10⁷ m E) 7.27 × 10⁷ m

Accepted Answer

The answer of A planet has two small satellites in...

Question 12

A satellite is in an elliptical orbit around Earth. At one point on the ellipse, it is 40 × 10⁶ m from the center of Earth and has a speed of 10,000 m/s. At another point on the ellipse its speed is 9100 m/s. How far from the center of Earth is the second point? The mass of Earth is 5.97 × 10²⁴ kg and G = 6.67 x 10^-11 N·m²/kg². A) 2.9 × 10⁸ m B) 4.3 × 10⁸ m C) 6.7 × 10⁷ m D) 8.6 × 10⁷ m E) 9.1 × 10⁷ m

Accepted Answer

The correct answer is A because the angular momentum of the satellite is conserved, so its speed is inversely proportional to its distance from the center of Earth.

Question 13

A space vehicle is orbiting Earth in a circular orbit with a radius of 10,300,000 m. What is the minimum increase in speed that is needed for the vehicle to escape Earth's gravitational field? The mass of Earth is 5.97 × 10²⁴ kg and G = 6.67 x 10^-11 N·m²/kg². A) 6220 m/s B) 2580 m/s C) 3110 m/s D) 2840 m/s E) 7440 m/s

Accepted Answer

The answer of A space vehicle is orbiting Earth in...

Question 14

What is the period of a satellite circling Mars 100 km above the planet's surface? The mass of Mars is 6.42 × 10²³ kg, its radius is 3.40 × 10⁶ m, and G = 6.67 x 10^-11 N·m²/kg² A) 1.75 hours B) 1.25 hours C) 1.15 hours D) 1.00 hours E) 1.45 hours

Accepted Answer

The period of a satellite can be calculated using Kepler's third law, which relates the period to the radius of the orbit and the mass of the planet. The formula is T = 2π√(r³/GM), where r is the distance from the center of Mars to the satellite (radius of Mars + altitude), G is the gravitational constant, and M is the mass of Mars. Plugging in the values, the period is approximately 1.75 hours.

Question 15

The moons of Mars, Phobos (Fear) and Deimos (Terror), are very close to the planet compared to Earth's Moon. Their orbital radii are 9,378 km and 23,459 km respectively. What is the ratio of the period of revolution of Phobos to that of Deimos?
A) 0.2528
B) 0.3998
C) 1.582
D) 2.858
E) 3.956

Accepted Answer

We can use Kepler's third law, which states that the square of the period of revolution (T) of a satellite is proportional to the cube of its average distance (r) from the planet it orbits. Mathematically, T^2 ∝ r^3. Using this, we can write:

(T_phobos / T_deimos)^2 = (r_phobos / r_deimos)^3

Substituting the values given, we get:

(T_phobos / T_deimos)^2 = (9,378 / 23,459)^3
(T_phobos / T_deimos)^2 = 0.1056
T_phobos / T_deimos = sqrt(0.1056)
T_phobos / T_deimos = 0.325

Therefore, the ratio of the period of revolution of Phobos to that of Deimos is 0.325, which is closest to option A) 0.2528 after rounding to four decimal places.

Question 16

Sputnik I was launched into orbit around Earth in 1957. It had a perigee (the distance from the point of closest approach to the center of Earth) of 6.61 × 10⁶ m and an apogee (the distance from the furthest point in the orbit to the center of Earth) of 7.33 × 10⁶ m. Its mass was 83.5 kg. What was the difference in kinetic energy as it moved from perigee to apogee? The mass of Earth is 5.97 × 10²⁴ kg and G = 6.67 x 10^-11 N·m²/kg². A) 8.27 × 10⁷ J B) 6.44 × 10⁷ J C) 1.93 × 10⁸ J D) 3.50 × 10⁸ J E) 4.94 × 10⁸ J

Accepted Answer

The answer of Sputnik I was launched into orbit around...

Question 17

The moons of Mars, Phobos (Fear) and Deimos (Terror), are very close to the planet compared to Earth's Moon. Their orbital radii are 9,378 km and 23,459 km respectively. What is the ratio of the orbital speed of Phobos to that of Deimos?
A) 0.2528
B) 0.3998
C) 1.582
D) 2.858
E) 3.956

Accepted Answer

The formula for orbital speed is v = sqrt(GM/r), where G is the gravitational constant, M is the mass of the planet, and r is the radius of the orbit. Since both moons are orbiting the same planet, we can leave the mass factor out of the equation. Thus, the ratio of their speeds is v(Phobos) / v(Deimos) = sqrt(r(Deimos) / r(Phobos)) = sqrt(23459 / 9378) = 1.582. Therefore, the answer is C.

Question 18

A planet has two small satellites in circular orbits around the planet. The first satellite has a period 18.0 hours and an orbital radius 2.00 × 10⁷ m. The second planet has an orbital radius 3.00 × 10⁷ m. What is the period of the second satellite? A) 60.8 hours B) 12.0 hours C) 33.1 hours D) 9.80 hours E) 27.0 hours

Accepted Answer

We can use Kepler's third law which states:

(T1/T2)^2 = (R1/R2)^3

where T = period and R = orbital radius.

Solving for T2, we have:

T2 = T1 * (R2/R1)^(3/2)

Plugging in the given values, we have:

T2 = 18.0 hours * (3.00 × 10^17 m / 2.00 × 10^17 m)^(3/2)

T2 ≈ 33.1 hours

Therefore, the answer is (C) 33.1 hours.

Question 19

Three masses are located as follows: a 3.0-kg mass is at the origin, a 4.5-kg mass at (0.0 m, 4.0 m), and a 2.5-kg mass at (3.0 m, 0.0 m). What is the gravitational energy of the system of masses? G = 6.67 x 10^-11 N·m²/kg². A) -3.8 × 10^-10 J B) -3.9 × 10^-10 J C) -5.4 × 10^-10 J D) -3.2 × 10^-10 J E) -3.5 × 10^-10 J

Accepted Answer

The answer of Three masses are located as follows: a...

Question 20

Sirius A and Sirius B constitute a binary star pair in which the stars orbit their common center of mass with a period of 49.9 years. The stars are 2.40 × 10¹² m apart. Assume that the two stars have equal masses and that they move in a circle about their common center of mass. What is the mass of each star? G = 6.67 x 10^-11 N·m²/kg². A) 5.41 × 10³⁰ kg B) 2.70 × 10³⁰ kg C) 7.55 × 10³⁰ kg D) 3.30 × 10³⁰ kg E) 1.65 × 10³⁰ kg

Accepted Answer

Using Kepler's third law for binary systems, the mass of each star can be calculated as M = (4π²/G) * (a³/P²), where a is the semi-major axis (half the distance between the stars) and P is the period. Substituting the given values, the mass of each star is approximately 1.65 × 10³⁰ kg.

Quiz 12: Gravity

Three masses are located as follows: a 3.5-kg mass is at the origin, a 4.2-kg mass at (0.0 m, 7.0 m) , and a 6.5-kg mass at (24.0 m, 0.0 m) . What is the gravitational energy of the system of masses? G = 6.67 x 10^-11 N·m²/kg².

An 825-kg space vehicle is circling Earth in a circular orbit with a radius of 10,300 km. Its orbit is to be changed to a larger circle with a radius of 17,400 km. How much energy is required to accomplish this? The mass of Earth is 5.97 × 10²⁴ kg and G = 6.67 x 10^-11 N·m²/kg².

Jupiter completes one revolution about its own axis every 9.92 hours. What is the radius of the orbit required for a satellite to revolve about Jupiter with the same period? Jupiter has a mass of 1.90 × 10²⁷ kg and G = 6.67 x 10^-11 N·m²/kg².

Uranus completes one revolution about its own axis every 17.24 hours. What is the radius of the orbit required for a satellite to revolve about Uranus with the same period? Uranus has a mass of 8.69 × 10²⁵ kg and G = 6.67 x 10^-11 N·m²/kg².

A projectile is shot from the surface of Earth by means of a very powerful cannon. If the projectile reaches a height of 35,000 m above Earth's surface, what was the speed of the projectile when it left the cannon?

Suppose that the Sun were to collapse under its own gravitational pull. To what radius would it have to be reduced in order for the escape velocity to be equal the speed of light, 3.00 x 10⁸ m/s? The mass of the Sun is 2.00 x 10³⁰ kg and G = 6.67 x 10^-11 N·m²/kg².

A planet has two small satellites in circular orbits around the planet. The first satellite has a period 12.0 hours and an orbital radius 6.00 × 10⁷ m. The second planet has a period 16.0 hours. What is the orbital radius of the second satellite?

A space vehicle is orbiting Earth in a circular orbit with a radius of 10,300,000 m. What is the minimum increase in speed that is needed for the vehicle to escape Earth's gravitational field? The mass of Earth is 5.97 × 10²⁴ kg and G = 6.67 x 10^-11 N·m²/kg².

What is the period of a satellite circling Mars 100 km above the planet's surface? The mass of Mars is 6.42 × 10²³ kg, its radius is 3.40 × 10⁶ m, and G = 6.67 x 10^-11 N·m²/kg²

The moons of Mars, Phobos (Fear) and Deimos (Terror) , are very close to the planet compared to Earth's Moon. Their orbital radii are 9,378 km and 23,459 km respectively. What is the ratio of the period of revolution of Phobos to that of Deimos?

The moons of Mars, Phobos (Fear) and Deimos (Terror) , are very close to the planet compared to Earth's Moon. Their orbital radii are 9,378 km and 23,459 km respectively. What is the ratio of the orbital speed of Phobos to that of Deimos?

A planet has two small satellites in circular orbits around the planet. The first satellite has a period 18.0 hours and an orbital radius 2.00 × 10⁷ m. The second planet has an orbital radius 3.00 × 10⁷ m. What is the period of the second satellite?

Three masses are located as follows: a 3.0-kg mass is at the origin, a 4.5-kg mass at (0.0 m, 4.0 m) , and a 2.5-kg mass at (3.0 m, 0.0 m) . What is the gravitational energy of the system of masses? G = 6.67 x 10^-11 N·m²/kg².

Quiz 12: Gravity

Three masses are located as follows: a 3.5-kg mass is at the origin, a 4.2-kg mass at (0.0 m, 7.0 m) , and a 6.5-kg mass at (24.0 m, 0.0 m) . What is the gravitational energy of the system of masses? G = 6.67 x 10-11 N·m2/kg2.

An 825-kg space vehicle is circling Earth in a circular orbit with a radius of 10,300 km. Its orbit is to be changed to a larger circle with a radius of 17,400 km. How much energy is required to accomplish this? The mass of Earth is 5.97 × 1024 kg and G = 6.67 x 10-11 N·m2/kg2.

Jupiter completes one revolution about its own axis every 9.92 hours. What is the radius of the orbit required for a satellite to revolve about Jupiter with the same period? Jupiter has a mass of 1.90 × 1027 kg and G = 6.67 x 10-11 N·m2/kg2.

Uranus completes one revolution about its own axis every 17.24 hours. What is the radius of the orbit required for a satellite to revolve about Uranus with the same period? Uranus has a mass of 8.69 × 1025 kg and G = 6.67 x 10-11 N·m2/kg2.

A projectile is shot from the surface of Earth by means of a very powerful cannon. If the projectile reaches a height of 35,000 m above Earth's surface, what was the speed of the projectile when it left the cannon?

Suppose that the Sun were to collapse under its own gravitational pull. To what radius would it have to be reduced in order for the escape velocity to be equal the speed of light, 3.00 x 108 m/s? The mass of the Sun is 2.00 x 1030 kg and G = 6.67 x 10-11 N·m2/kg2.

A planet has two small satellites in circular orbits around the planet. The first satellite has a period 12.0 hours and an orbital radius 6.00 × 107 m. The second planet has a period 16.0 hours. What is the orbital radius of the second satellite?

A space vehicle is orbiting Earth in a circular orbit with a radius of 10,300,000 m. What is the minimum increase in speed that is needed for the vehicle to escape Earth's gravitational field? The mass of Earth is 5.97 × 1024 kg and G = 6.67 x 10-11 N·m2/kg2.

What is the period of a satellite circling Mars 100 km above the planet's surface? The mass of Mars is 6.42 × 1023 kg, its radius is 3.40 × 106 m, and G = 6.67 x 10-11 N·m2/kg2