Gravitation — Set 3
Physics · गुरुत्वाकर्षण · Questions 21–30 of 60
Two masses 2 kg and 3 kg are 2 m apart. Using G = 6.67×10^-11 N m^2 kg^-2, what is the gravitational force between them (approximately)?
Correct Answer: C. 1.0×10^-10 N
Use F = Gm1m2/r^2 = 6.67×10^-11×(2×3)/2^2. That is 6.67×10^-11×6/4 ≈ 1.0×10^-10 N. The force is extremely small because G is very small.
At a distance 2R from Earth’s center (R = Earth’s radius), the value of g compared to its surface value g is?
Correct Answer: C. g/4
Gravity varies as 1/r^2 from the center. At r = 2R, g' = g(R^2/(2R)^2) = g/4. So it becomes one-fourth of the surface value.
Using v = sqrt(gR) with g = 9.8 m/s^2 and R = 6.4×10^6 m, the orbital speed near Earth’s surface is closest to?
Correct Answer: B. 7.9 km/s
Compute v = sqrt(gR) = sqrt(9.8×6.4×10^6). This is sqrt(6.272×10^7) ≈ 7.92×10^3 m/s = 7.9 km/s. This is also called the first cosmic speed near Earth.
Using v_escape = sqrt(2gR) with g = 9.8 m/s^2 and R = 6.4×10^6 m, the escape velocity from Earth is closest to?
Correct Answer: D. 11.2 km/s
Compute v = sqrt(2gR) = sqrt(2×9.8×6.4×10^6). This equals sqrt(1.2544×10^8) ≈ 1.12×10^4 m/s = 11.2 km/s. Escape speed is √2 times the near-surface orbital speed.
If the orbital radius of a satellite is increased by a factor of 8, its time period changes by what factor?
Correct Answer: C. 8^(3/2) ≈ 22.6 times
Kepler’s third law gives T ∝ r^(3/2). If r becomes 8r, then T becomes 8^(3/2)T. Since 8^(3/2) = sqrt(8^3) = sqrt(512) ≈ 22.6, the period increases about 22.6 times.
Why is a person’s weight generally slightly less at the equator than at the poles?
Correct Answer: B. Because effective g is reduced by Earth’s rotation and larger radius at the equator
At the equator, Earth’s rotation provides a small outward centrifugal effect. Also the equatorial radius is slightly larger, increasing distance from the center. Both effects reduce effective g and thus reduce weight.
The magnitude of gravitational field strength at distance r from a point mass M is?
Correct Answer: C. GM/r^2
Gravitational field strength is force per unit mass. For a point mass, F = GMm/r^2, so g = F/m = GM/r^2. It points toward the mass.
The principle of superposition in gravitation states that?
Correct Answer: D. Net gravitational force equals the vector sum of forces due to individual masses
Each mass produces its own gravitational force independently. The total force on a body is the vector sum of all these forces. This is called the superposition principle.
For a spherically symmetric body, the gravitational field outside it behaves as if?
Correct Answer: C. All its mass were concentrated at its center
Outside a spherical mass distribution, gravity depends only on total mass. The field is the same as that of a point mass at the center. This result follows from spherical symmetry.
The first cosmic velocity for a circular orbit close to Earth’s surface is?
Correct Answer: C. sqrt(gR)
First cosmic velocity is the minimum speed for a low circular orbit. It is v = sqrt(GM/R), which equals sqrt(gR). It is about 7.9 km/s for Earth.