Question 1

Refer to Exhibit 11-1 below.Exhibit 11-1 Three 5.0-kg masses are located at points in the xy plane as shown in the figure.  What is the magnitude of the resultant force (caused by the other two masses) on the mass at the origin?
A)2.7 * 10-8 N
B)2.1 *10-8 N
C)1.8 * 10-8 N
D)2.4 * 10-8 N
E)2.9 * 10-8 N

Accepted Answer

B

Question 2

What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is the radius of the Earth?
A)4.8 m/s2
B)1.1 m/s2
C)3.3 m/s2
D)2.5 m/s2
E)6.5 m/s2

Accepted Answer

The gravitational acceleration at a distance from the Earth's center is given by $ g' = \frac{g}{(1 + h/R)^2} $, where $ h $ is the height above the Earth's surface and $ R $ is the Earth's radius. At 2R above the surface, $ h = 2R $, so $ g' = \frac{9.8}{(1 + 2)^2} = \frac{9.8}{9} \approx 1.1 \, \text{m/s}^2 $.

Question 3

Refer to Exhibit 11-1 below.Exhibit 11-1 Three 5.0-kg masses are located at points in the xy plane as shown in the figure.  What is the magnitude of the resultant force (caused by the other two masses) on the mass at x = 0.40 m, y = 0?
A)2.2 * 10-8 N
B)1.9 *10-8 N
C)1.4 * 10-8 N
D)1.6 * 10-8 N
E)2.5 * 10-8 N

Accepted Answer

To find the resultant force on the mass at x = 0.4 m, y=0, we need to find the vector sum of the forces exerted on it by the other two masses. Using the formula for gravitational force between two masses ($F=G\frac{m_1m_2}{r^2}$) and the Pythagorean theorem, we can find the magnitudes of the forces exerted by the two masses as follows:
- Magnitude of force exerted by the 2.0-kg mass at (-0.60 m, 0):
$F_1 = G\frac{m_1m_2}{r^2} = (6.67	imes10^{-11} Nm^2 / kg^2) \frac{(2.0 kg)(5.0 kg)}{(-0.60 m)^2} = 1.11	imes10^{-9} N$
- Magnitude of force exerted by the 3.0-kg mass at (0.60 m, 0):
$F_2 = G\frac{m_1m_2}{r^2} = (6.67	imes10^{-11} Nm^2 / kg^2) \frac{(3.0 kg)(5.0 kg)}{(0.60 m)^2} = 3.70	imes10^{-9} N$
These forces are both attractive (towards the mass at x=0.4 m, y=0), so we can add them together as vectors to find the net force. To do this, we need to find the x and y components of each force:
- $F_{1x} = -F_1\sin{	heta_1} = -(1.11	imes10^{-9} N)\frac{0.60 m}{\sqrt{(0.60 m)^2 + (0.40 m)^2}} = -5.56 	imes 10^{-10} N$
- $F_{1y} = F_1\cos{	heta_1} = (1.11	imes10^{-9} N)\frac{0.40 m}{\sqrt{(0.60 m)^2 + (0.40 m)^2}} = 3.70	imes10^{-10} N$
- $F_{2x} = F_2\sin{	heta_2} = (3.70	imes10^{-9} N)\frac{0.60 m}{\sqrt{(0.60 m)^2 + (0.40 m)^2}} = 1.85	imes10^{-9} N$
- $F_{2y} = F_2\cos{	heta_2} = (3.70	imes10^{-9} N)\frac{0.40 m}{\sqrt{(0.60 m)^2 + (0.40 m)^2}} = 1.23	imes10^{-9} N$
where $	heta_1$ and $	heta_2$ are the angles that the forces make with the x-axis.
Now we can add the x and y components of the forces to find the net force:
$F_{net,x} = F_{1x} + F_{2x} = -5.56 	imes 10^{-10} N + 1.85	imes10^{-9} N = 1.29	imes10^{-9} N$
$F_{net,y} = F_{1y} + F_{2y} = 3.70	imes10^{-10} N + 1.23	imes10^{-9} N = 1.60	imes10^{-9} N$
The magnitude of the net force is then:
$F_{net} = \sqrt{F_{net,x}^2 + F_{net,y}^2} = \sqrt{(1.29	imes10^{-9} N)^2 + (1.60	imes10^{-9} N)^2} = 1.6	imes10^{-9} N \approx \boxed{	extbf{(D)}\ 1.6	imes10^{-9} N}$

Question 4

Three 5.0-kg masses are located at points in the xy plane, as shown. What is the magnitude of the resultant force (caused by the other two masses) on the mass at x = 0, y = 0.30 m? 
A)2.6 * 10-8 N
B)2.0 *10-8 N
C)2.9 * 10-8 N
D)2.3 * 10-8 N
E)2.1 * 10-8 N

Accepted Answer

The answer of Three 5.0-kg masses are located at points...

Question 5

An object is released from rest at a distance h above the surface of a planet (mass = M, radius = R < h). With what speed will the object strike the surface of the planet? Disregard any dissipative effects of the atmosphere of the planet.
A) $\left[\frac{2 G M h}{R(R+h)}\right]^{1 / 2}$
B) $\left[\frac{2 G M M}{R}\right]^{1 / 2}$
C) $\left[\frac{2 G M(h-R)}{R h}\right]^{1 / 2}$
D) $\left[\frac{2 G M M}{R+h}\right]^{1 / 2}$
E) $\left[\frac{2 G M /}{R-h}\right]^{1 / 2}$

Accepted Answer

The answer of An object is released from rest at...

Question 6

A satellite circles planet Roton every 2.8 h in an orbit having a radius of 1.2* 107 m. If the radius of Roton is 5.0* 106 m, what is the magnitude of the free-fall acceleration on the surface of Roton?
A)31 m/s2
B)27 m/s2
C)34 m/s2
D)40 m/s2
E)19 m/s2

Accepted Answer

The gravitational force provides the centripetal force for the satellite. Using the formula for centripetal acceleration and the gravitational force, we can solve for the gravitational acceleration at the surface of Roton. The formula is g = (4π²r³)/(T²R²), where r is the orbit radius, T is the period, and R is the planet's radius. Plugging in the values gives approximately 27 m/s².

Question 7

An object is released from rest when it is a height h above the surface of a planet of mass M and radius R. What is the speed of the object just before striking the surface of the planet? Neglect any air resistance. Let h = 4.0 * 106 m, R = 5.0 * 106 m, and M = 4.0 * 1024 kg.
A)7.8 km/s
B)3.5 km/s
C)5.4 km/s
D)6.9 km/s
E)4.8 km/s

Accepted Answer

Using conservation of energy:
Initial energy = mgh
Final energy = 1/2 mv^2 - GMm/R
Setting them equal and solving for v:
v = sqrt(2GM/R + 2gh) = 6.9 km/s (rounded to one decimal place)

Question 8

What is the escape speed from a planet of mass M and radius R if M = 3.2 * 1023 kg and R = 2.4 * 106 m?
A)5.5 km/s
B)4.2 km/s
C)5.2 km/s
D)4.8 km/s
E)3.7 km/s

Accepted Answer

The escape speed is calculated using the formula v = sqrt(2GM/R). Plugging in the values G = 6.674 * 10^-11 m^3/kg/s^2, M = 3.2 * 10^23 kg, and R = 2.4 * 10^6 m, the escape speed is approximately 4.2 km/s.

Question 9

A projectile is launched from the surface of a planet (mass = M, radius = R). What minimum launch speed is required if the projectile is to rise to a height of 2R above the surface of the planet? Disregard any dissipative effects of the atmosphere.
A) $\left[\frac{4 G M R}{3 R}\right]^{1 / 2}$
B) $\left[\frac{8 G M R}{5 R}\right]^{1 / 2}$
C) $\left[\frac{3 G M R}{2 R}\right]^{1 / 2}$
D) $\left[\frac{5 G M R}{3 R}\right]^{1 / 2}$
E) $\left[\frac{G M}{3 R}\right]^{1 / 2}$

Accepted Answer

The answer of A projectile is launched from the surface...

Question 10

Planet Zero has a mass of 5.0 *1023 kg and a radius of 2.0 * 106 m. A space probe is launched vertically from the surface of Zero with an initial speed of 4.0 km/s. What is the speed of the probe when it is 3.0 *06 m from Zero's centre?
A)3.0 km/s
B)2.2 km/s
C)1.6 km/s
D)3.7 km/s
E)5.9 km/s

Accepted Answer

The speed of the probe can be found using the conservation of energy principle. The initial kinetic energy plus the initial gravitational potential energy equals the final kinetic energy plus the final gravitational potential energy. Solving this equation gives the final speed as 2.2 km/s.

Question 11

The period of a satellite circling planet Nutron is observed to be 84 s when it is in a circular orbit with a radius of 8.0 * 106 m. What is the mass of planet Nutron?
A)6.2 * 1028 kg
B)5.0 *1028 kg
C)5.5 * 1028 kg
D)4.3 * 1028 kg
E)3.7 * 1028 kg

Accepted Answer

The answer of The period of a satellite circling planet...

Question 12

A 50-kg satellite circles planet Cruton every 5.6 h in an orbit with a radius of 12 * 106 m. What is the magnitude of the gravitational force on the satellite by planet Cruton?
A)63 N
B)58 N
C)68 N
D)73 N
E)50 N

Accepted Answer

Using the formula for gravitational force:

$F = \frac{Gm_1m_2}{r^2}$

where 
- $G$ is the gravitational constant ($6.67	imes10^{-11}$ N$m^2/kg^2$), 
- $m_1$ and $m_2$ are the masses of the two objects, and 
- $r$ is the distance between their centers.

Converting the given values into SI units:

$m_1 =$ mass of planet Cruton $= ?$ (not given) 
$m_2 =$ mass of satellite $= 50$ kg 
$r =$ radius of satellite's orbit $= 12	imes10^6$ m

We can use Kepler's third law to find the mass of planet Cruton:

$T^2 = \frac{4\pi^2}{GM}r^3$

where 
- $T$ is the period of the satellite's orbit ($5.6$ hours $= 20,160$ seconds), 
- $G$ is the gravitational constant, 
- $M$ is the mass of the planet.

Solving for $M$:

$M = \frac{4\pi^2}{GT^2}r^3 = \frac{4\pi^2}{(6.67	imes10^{-11})(20,160)^2}(12	imes10^6)^3 = 6.12	imes10^{24}$ kg

Now we can calculate the gravitational force:

$F = \frac{Gm_1m_2}{r^2} = \frac{(6.67	imes10^{-11})(6.12	imes10^{24})(50)}{(12	imes10^6)^2} \approx 58	ext{ N}$

Therefore, the answer is B) 58 N.

Question 13

A spacecraft (mass = m) orbits a planet (mass = M) in a circular orbit (radius = R). What is the minimum energy required to send this spacecraft to a distant point in space where the gravitational force on the spacecraft by the planet is negligible?
A)GmM/(4R)
B)GmM/R
C)GmM/(2R)
D)GmM/(3R)
E)2GmM/(5R)

Accepted Answer

The minimum energy required to send the spacecraft to a distant point in space where the gravitational force on the spacecraft by the planet is negligible is equal to the energy required to overcome the gravitational potential energy at that point. At infinity, the gravitational potential energy is zero. Therefore, the energy required is the difference between the initial energy, Ei, and the potential energy at infinity, Ef:

E = Ei - Ef

The initial energy of the spacecraft in circular orbit is given by:

Ei = -GmM/(2R)

To find the potential energy at infinity, we use the fact that the potential energy is given by:

U = -GmM/r

where r is the distance between the spacecraft and the planet. At infinity, r approaches infinity, therefore the potential energy at infinity is zero:

Ef = 0

Substituting the values of Ei and Ef, we get:

E = -GmM/(2R) - 0 = -GmM/(2R)

Therefore, the minimum energy required to send the spacecraft to a distant point in space where the gravitational force on the spacecraft by the planet is negligible is GmM/(2R). The answer is choice C.

Question 14

What is the gravitational force on a 20-kg satellite circling the Earth (radius = 6.4 * 106 m, mass = 6.0 * 1024 kg) with a period of 5.0 h?
A)88 N
B)55 N
C)36 N
D)98 N
E)18 N

Accepted Answer

The answer of What is the gravitational force on a...

Question 15

A spaceship of mass m circles a planet (mass = M) in an orbit of radius R. How much energy is required to transfer the spaceship to a circular orbit of radius 3R?
A)GmM/(2R)
B)GmM/(3R)
C)GmM/(4R)
D)GmM/(6R)
E)3GmM/(4R)

Accepted Answer

The answer of A spaceship of mass m circles a...

Question 16

Planet Roton has a mass of 4.0* 1023 kg and a radius of 2.0 * 106 m. With what speed should a space probe be launched from the surface of Roton so as to achieve a maximum distance of 3.0 * 106 m from the centre of Roton?
A)4.2 km/s
B)3.9 km/s
C)3.0 km/s
D)3.4 km/s
E)6.0 km/s

Accepted Answer

The answer of Planet Roton has a mass of 4.0*...

Question 17

What is the kinetic energy of a 200-kg satellite as it follows a circular orbit of radius 8.0 *106 m around the Earth? (Mass of Earth = 6.0 * 1024 kg.)
A)5.0 *109 J
B)1.0 * 1010 J
C)1.5 * 1010 J
D)2.0 * 1010 J
E)2.5 * 109 J

Accepted Answer

Kinetic energy = 1/2 mv^2 = 1/2 * 200 kg * (v)^2v = sqrt(GM/r) = sqrt((6.67 * 10^-11 N m^2 / kg^2 * 6.0 * 10^24 kg) / (8.0 * 10^6 m)) = 7900 m/sKinetic energy = 1/2 * 200 kg * (7900 m/s)^2 = 5.0 * 10^9 J

Question 18

A satellite is in a circular orbit about the Earth at an altitude at which air resistance is negligible. Which of the following statements is true?
A)There is only one force acting on the satellite.
B)There are two forces acting on the satellite, and their resultant is zero.
C)There are two forces acting on the satellite, and their resultant is not zero.
D)There are three forces acting on the satellite.
E)None of the preceding statements are correct.

Accepted Answer

According to Newton's First Law of Motion, an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. In this case, the satellite is in circular motion, which means that it is being constantly pulled towards the center of the Earth by the force of gravity. Since air resistance is negligible at this altitude, there are no other external forces acting on the satellite. Therefore, the only force acting on the satellite is the force of gravity.

Question 19

A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to another circular orbit with a period of 180 min? (Earth: radius = 6.4 * 106 m, mass = 6.0 * 1024 kg)
A)2.9 *108 J
B)3.5 * 108 J
C)4.1 * 108 J
D)4.7* 108 J
E)5.9 * 108 J

Accepted Answer

The minimum energy required to change the orbit is equal to the difference in the potential energies of the two orbits.

Let r1 and r2 be the radii of the two orbits, T1 and T2 be their periods, and E be the energy required to change the orbit.

Using Kepler's third law, we can write:

T1^2/r1^3 = T2^2/r2^3

Solving for r2, we get:

r2 = (T2/T1)^(2/3) * r1

We are given r1, T1, and T2, so we can calculate r2:

r2 = (180/120)^(2/3) * 6.4 x 10^6 m = 8.0 x 10^6 m

The potential energy of a satellite in circular orbit is given by:

U = -G*M*m/r

where G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the radius of the orbit.

The potential energy of the satellite in its initial orbit is:

U1 = -G*M*m/r1

The potential energy of the satellite in its final orbit is:

U2 = -G*M*m/r2

The minimum energy required to change the orbit is:

E = U2 - U1 = G*M*m*(1/r1 - 1/r2)

Substituting the values given, we get:

E = 6.67 x 10^-11 N*m^2/kg^2 * 6.0 x 10^24 kg * 50 kg * (1/6.4 x 10^6 m - 1/8.0 x 10^6 m)
E = 2.9 x 10^8 J

Therefore, the correct answer is (A).

Question 20

Two stars of masses M and 6M are separated by a distance D. Determine the distance (measured from M) to a point at which the net gravitational force on a third mass would be zero.
A)0.41 D
B)0.33 D
C)0.37 D
D)0.29 D
E)0.14 D

Accepted Answer

The answer of Two stars of masses M and 6M...

Quiz 11: Gravity

Refer to Exhibit 11-1 below.Exhibit 11-1 Three 5.0-kg masses are located at points in the xy plane as shown in the figure. What is the magnitude of the resultant force (caused by the other two masses) on the mass at the origin?

What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is the radius of the Earth?

Refer to Exhibit 11-1 below.Exhibit 11-1 Three 5.0-kg masses are located at points in the xy plane as shown in the figure. What is the magnitude of the resultant force (caused by the other two masses) on the mass at x = 0.40 m, y = 0?

Three 5.0-kg masses are located at points in the xy plane, as shown. What is the magnitude of the resultant force (caused by the other two masses) on the mass at x = 0, y = 0.30 m?

An object is released from rest at a distance h above the surface of a planet (mass = M, radius = R < h) . With what speed will the object strike the surface of the planet? Disregard any dissipative effects of the atmosphere of the planet.

A satellite circles planet Roton every 2.8 h in an orbit having a radius of 1.2* 107 m. If the radius of Roton is 5.0* 106 m, what is the magnitude of the free-fall acceleration on the surface of Roton?

An object is released from rest when it is a height h above the surface of a planet of mass M and radius R. What is the speed of the object just before striking the surface of the planet? Neglect any air resistance. Let h = 4.0 * 106 m, R = 5.0 * 106 m, and M = 4.0 * 1024 kg.

What is the escape speed from a planet of mass M and radius R if M = 3.2 * 1023 kg and R = 2.4 * 106 m?

A projectile is launched from the surface of a planet (mass = M, radius = R) . What minimum launch speed is required if the projectile is to rise to a height of 2R above the surface of the planet? Disregard any dissipative effects of the atmosphere.

Planet Zero has a mass of 5.0 1023 kg and a radius of 2.0 106 m. A space probe is launched vertically from the surface of Zero with an initial speed of 4.0 km/s. What is the speed of the probe when it is 3.0 *06 m from Zero's centre?

The period of a satellite circling planet Nutron is observed to be 84 s when it is in a circular orbit with a radius of 8.0 * 106 m. What is the mass of planet Nutron?

A 50-kg satellite circles planet Cruton every 5.6 h in an orbit with a radius of 12 * 106 m. What is the magnitude of the gravitational force on the satellite by planet Cruton?

A spacecraft (mass = m) orbits a planet (mass = M) in a circular orbit (radius = R) . What is the minimum energy required to send this spacecraft to a distant point in space where the gravitational force on the spacecraft by the planet is negligible?

What is the gravitational force on a 20-kg satellite circling the Earth (radius = 6.4 * 106 m, mass = 6.0 * 1024 kg) with a period of 5.0 h?

A spaceship of mass m circles a planet (mass = M) in an orbit of radius R. How much energy is required to transfer the spaceship to a circular orbit of radius 3R?

Planet Roton has a mass of 4.0* 1023 kg and a radius of 2.0 * 106 m. With what speed should a space probe be launched from the surface of Roton so as to achieve a maximum distance of 3.0 * 106 m from the centre of Roton?

What is the kinetic energy of a 200-kg satellite as it follows a circular orbit of radius 8.0 106 m around the Earth? (Mass of Earth = 6.0 1024 kg.)

A satellite is in a circular orbit about the Earth at an altitude at which air resistance is negligible. Which of the following statements is true?

A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to another circular orbit with a period of 180 min? (Earth: radius = 6.4 * 106 m, mass = 6.0 * 1024 kg)

Two stars of masses M and 6M are separated by a distance D. Determine the distance (measured from M) to a point at which the net gravitational force on a third mass would be zero.