Gravitation
Q. What is the gravitational potential energy of a mass of 10 kg at a height of 5 m above the ground? (2000)
A.
500 J
B.
1000 J
C.
1500 J
D.
2000 J
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Solution
Gravitational potential energy U = mgh = 10 kg * 9.8 m/s² * 5 m = 490 J, approximately 500 J.
Correct Answer: B — 1000 J
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Q. What is the gravitational potential energy of a mass of 10 kg at a height of 5 m above the ground? (g = 9.8 m/s²)
A.
490 J
B.
98 J
C.
588 J
D.
0 J
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Solution
U = mgh = 10 * 9.8 * 5 = 490 J
Correct Answer: A — 490 J
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Q. What is the gravitational potential energy of a mass of 10 kg at a height of 5 m in a gravitational field of strength 9.8 N/kg?
A.
490 J
B.
50 J
C.
98 J
D.
0 J
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Solution
Gravitational potential energy U = mgh = 10 * 9.8 * 5 = 490 J.
Correct Answer: A — 490 J
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Q. What is the gravitational potential energy of a mass of 2 kg at a height of 10 m? (g = 9.8 m/s²)
A.
19.6 J
B.
39.2 J
C.
78.4 J
D.
98 J
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Solution
PE = mgh = 2 * 9.8 * 10 = 196 J
Correct Answer: B — 39.2 J
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Q. What is the gravitational potential energy of a mass of 2 kg at a height of 10 m above the ground? (g = 9.8 m/s^2)
A.
196 J
B.
98 J
C.
20 J
D.
10 J
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Solution
Gravitational potential energy (U) = mgh = 2 kg * 9.8 m/s^2 * 10 m = 196 J.
Correct Answer: A — 196 J
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Q. What is the gravitational potential energy of a mass of 2 kg at a height of 10 m in a gravitational field of 9.8 m/s²?
A.
19.6 J
B.
39.2 J
C.
78.4 J
D.
98 J
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Solution
Gravitational potential energy (U) = mgh = 2 kg * 9.8 m/s² * 10 m = 196 J.
Correct Answer: B — 39.2 J
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Q. What is the gravitational potential energy of a satellite of mass m at a height h above the Earth's surface?
A.
-GMm/R
B.
-GMm/(R+h)
C.
-GMm/(R-h)
D.
-GMm/(R^2 + h^2)
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Solution
The gravitational potential energy of a satellite at height h is given by U = -GMm/(R+h), where R is the radius of the Earth.
Correct Answer: B — -GMm/(R+h)
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Q. What is the gravitational potential energy of an object of mass m at a height h above the Earth's surface?
A.
mgh
B.
gh/m
C.
mg/h
D.
mgh^2
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Solution
The gravitational potential energy is given by the formula U = mgh.
Correct Answer: A — mgh
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Q. What is the gravitational potential energy of two masses m1 and m2 separated by a distance r?
A.
-G * (m1 * m2) / r
B.
G * (m1 * m2) / r
C.
G * (m1 + m2) / r
D.
-G * (m1 + m2) / r
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Solution
The gravitational potential energy is given by U = -G * (m1 * m2) / r.
Correct Answer: A — -G * (m1 * m2) / r
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Q. What is the gravitational potential energy U of two masses m1 and m2 separated by a distance r?
A.
U = -G * (m1 * m2) / r
B.
U = G * (m1 * m2) / r
C.
U = (m1 + m2) * r
D.
U = G * (m1 - m2) / r
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Solution
The gravitational potential energy is given by U = -G * (m1 * m2) / r.
Correct Answer: A — U = -G * (m1 * m2) / r
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Q. What is the minimum speed required for a satellite to achieve a low Earth orbit?
A.
7.9 km/s
B.
11.2 km/s
C.
5.0 km/s
D.
9.8 km/s
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Solution
The minimum speed required for a satellite to achieve a low Earth orbit is approximately 7.9 km/s.
Correct Answer: A — 7.9 km/s
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Q. What is the minimum speed required for a satellite to maintain a low Earth orbit?
A.
7.9 km/s
B.
11.2 km/s
C.
5.0 km/s
D.
9.8 km/s
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Solution
The minimum speed required for a satellite in low Earth orbit is approximately 7.9 km/s.
Correct Answer: A — 7.9 km/s
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Q. What is the orbital speed of a satellite in a circular orbit at a height h above the Earth's surface?
A.
sqrt(GM/R^2)
B.
sqrt(GM/(R+h)^2)
C.
sqrt(GM/(R-h)^2)
D.
sqrt(GM/R)
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Solution
The orbital speed v of a satellite in a circular orbit is given by v = sqrt(GM/(R+h)^2), where G is the gravitational constant, M is the mass of the Earth, R is the radius of the Earth, and h is the height of the satellite above the Earth's surface.
Correct Answer: B — sqrt(GM/(R+h)^2)
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Q. What is the orbital speed of a satellite in a low Earth orbit (LEO) approximately 2000 km above the Earth's surface? (2000)
A.
7.9 km/s
B.
11.2 km/s
C.
5.0 km/s
D.
3.0 km/s
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Solution
The orbital speed of a satellite in low Earth orbit is approximately 7.9 km/s.
Correct Answer: A — 7.9 km/s
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Q. What is the period of a satellite in a circular orbit at a height of 300 km above the Earth's surface?
A.
90 minutes
B.
60 minutes
C.
120 minutes
D.
30 minutes
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Solution
The period of a satellite in a circular orbit at a height of 300 km is approximately 90 minutes.
Correct Answer: A — 90 minutes
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Q. What is the period of a satellite in a low Earth orbit (LEO) compared to a satellite in a geostationary orbit?
A.
Longer than a geostationary orbit
B.
Shorter than a geostationary orbit
C.
Equal to a geostationary orbit
D.
Depends on the mass of the satellite
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Solution
Satellites in low Earth orbit have a much shorter orbital period compared to geostationary satellites due to their proximity to Earth.
Correct Answer: B — Shorter than a geostationary orbit
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Q. What is the primary force acting on a satellite in a stable orbit around the Earth?
A.
Gravitational force
B.
Electromagnetic force
C.
Frictional force
D.
Centrifugal force
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Solution
A satellite in a stable orbit is primarily influenced by the gravitational force exerted by the Earth.
Correct Answer: A — Gravitational force
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Q. What is the relationship between gravitational field strength and gravitational potential?
A.
Field strength is the gradient of potential.
B.
Field strength is the integral of potential.
C.
They are independent.
D.
Field strength is the square of potential.
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Solution
Gravitational field strength is the negative gradient of gravitational potential.
Correct Answer: A — Field strength is the gradient of potential.
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Q. What is the relationship between gravitational potential and gravitational field strength?
A.
V = -g * r
B.
g = -dV/dr
C.
V = g * r
D.
g = dV/dr
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Solution
The gravitational field strength g is the negative gradient of the gravitational potential V, given by g = -dV/dr.
Correct Answer: B — g = -dV/dr
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Q. What is the relationship between the height of a satellite and its orbital period?
A.
Directly proportional
B.
Inversely proportional
C.
No relationship
D.
Exponential relationship
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Solution
The orbital period T of a satellite is related to its height h by T ∝ h^(3/2), which indicates that the period is inversely proportional to the square root of the gravitational force acting on it.
Correct Answer: B — Inversely proportional
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Q. What is the relationship between the orbital radius and the period of a satellite in a circular orbit?
A.
T is directly proportional to r
B.
T is inversely proportional to r
C.
T is proportional to r^2
D.
T is proportional to √r
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Solution
The period T of a satellite is proportional to the square root of the orbital radius r (T ∝ √r).
Correct Answer: D — T is proportional to √r
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Q. What is the relationship between the orbital radius and the time period of a satellite?
A.
T ∝ r^2
B.
T ∝ r^3/2
C.
T ∝ r
D.
T ∝ r^1/2
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Solution
The time period T of a satellite is related to the orbital radius r by T ∝ r^(3/2), according to Kepler's third law.
Correct Answer: B — T ∝ r^3/2
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Q. What is the relationship between the period of a satellite and its orbital radius?
A.
T is directly proportional to r
B.
T is inversely proportional to r
C.
T is proportional to r^2
D.
T is proportional to √r
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Solution
The period T of a satellite is proportional to the square root of the orbital radius r, as given by T = 2π√(r^3/GM).
Correct Answer: D — T is proportional to √r
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Q. What is the shape of the equipotential surfaces around a point mass?
A.
Straight lines
B.
Parabolas
C.
Spheres
D.
Cylinders
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Solution
The equipotential surfaces around a point mass are spherical.
Correct Answer: C — Spheres
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Q. What is the shape of the gravitational field lines around a point mass?
A.
Straight lines
B.
Curved lines
C.
Concentric circles
D.
Radial lines
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Solution
The gravitational field lines around a point mass are radial, pointing towards the mass.
Correct Answer: D — Radial lines
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Q. What is the unit of gravitational potential?
A.
J/kg
B.
N/kg
C.
J
D.
kg
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Solution
The unit of gravitational potential is Joules per kilogram (J/kg).
Correct Answer: A — J/kg
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Q. What is the value of gravitational acceleration (g) at the surface of the Earth?
A.
9.8 m/s²
B.
10 m/s²
C.
9.81 m/s²
D.
8.9 m/s²
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Solution
The standard value of gravitational acceleration at the Earth's surface is approximately 9.81 m/s².
Correct Answer: C — 9.81 m/s²
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Q. What is the value of gravitational acceleration at the height equal to the radius of the Earth?
A.
g/2
B.
g/4
C.
g/3
D.
g/8
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Solution
At height R, g' = g / (2^2) = g / 4.
Correct Answer: B — g/4
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Q. What is the value of gravitational acceleration at the surface of the Earth?
A.
9.8 m/s^2
B.
10 m/s^2
C.
9.81 m/s^2
D.
8.9 m/s^2
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Solution
The standard value of gravitational acceleration at the surface of the Earth is approximately 9.81 m/s^2.
Correct Answer: C — 9.81 m/s^2
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Q. What is the value of the gravitational constant G?
A.
6.67 x 10^-11 N(m/kg)^2
B.
9.81 m/s^2
C.
3.00 x 10^8 m/s
D.
1.60 x 10^-19 C
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Solution
The gravitational constant G is approximately 6.67 x 10^-11 N(m/kg)^2.
Correct Answer: A — 6.67 x 10^-11 N(m/kg)^2
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