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
• **F ≈ 1.0×10⁻¹⁰ N** = F = Gm₁m₂/r² = 6.67×10⁻¹¹ × (2×3)/2² = 6.67×10⁻¹¹ × 1.5 ≈ 1.0×10⁻¹⁰ N. • **G = 6.67×10⁻¹¹ N m² kg⁻²** — The tiny value of G makes gravitational force negligible between everyday objects. • 💡 Wrong-option analysis: 1.0×10⁻¹¹ N: ten times too small — missing the factor 1.5; 5.0×10⁻¹¹ N: about half the correct value; 2.0×10⁻¹⁰ N: twice the correct value, possibly using r = 1 m instead of r² = 4 m².
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
• **g/4** = Since g ∝ 1/r² from the center, at r = 2R: g′ = g × (R/2R)² = g/4. • **Rapid decrease with distance** — At 3R, g′ = g/9; at 4R, g′ = g/16. • 💡 Wrong-option analysis: 4g: would require being at half the radius, not double; 2g: confuses halving distance with doubling — inverse-square gives quartering; g/2: correct only if g ∝ 1/r rather than 1/r².
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
• **≈ 7.9 km/s** = v = √(gR) = √(9.8 × 6.4×10⁶) ≈ 7,919 m/s ≈ 7.9 km/s. • **First cosmic velocity** — The minimum speed for a circular orbit just above Earth's surface. • 💡 Wrong-option analysis: 5.6 km/s: about √2 times too small; 2.8 km/s: roughly 2.8 times too small; 11.2 km/s: that is escape velocity = √(2gR), not the orbital speed.
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
• **≈ 11.2 km/s** = v_e = √(2gR) = √(2 × 9.8 × 6.4×10⁶) ≈ 11,200 m/s. • **= √2 × orbital speed** — Escape velocity is exactly √2 times the circular orbital speed near Earth's surface. • 💡 Wrong-option analysis: 7.9 km/s: orbital (first cosmic) speed, not escape speed; 9.4 km/s: no standard physical significance; 10.0 km/s: does not match the √(2gR) calculation.
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
• **8^(3/2) ≈ 22.6 times** = By Kepler's third law T ∝ r^(3/2): new T = T × 8^(3/2) = T√512 ≈ 22.6T. • **T ∝ r^(3/2)** — Doubling orbital radius multiplies period by 2^1.5 ≈ 2.83. • 💡 Wrong-option analysis: √8 times: would apply if T ∝ r^(1/2), which is not Kepler's law; 1/8 times: period always increases with radius, never decreases; 8 times: applies only if T ∝ r (linear), not the r^(3/2) Kepler relation.
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
• **Reduced effective g at equator** = Earth bulges at equator (larger r → smaller g) and Earth's rotation causes a centrifugal correction that reduces apparent weight. • **Difference ≈ 0.5%** — Weight at equator is about 0.3–0.5% less than at the poles. • 💡 Wrong-option analysis: Earth's mass smaller at equator: total mass is the same; Air pressure higher at equator: atmospheric pressure does not directly change gravitational force; Gravity zero at equator: g ≈ 9.78 m/s² at equator, far from zero.
The magnitude of gravitational field strength at distance r from a point mass M is?
Correct Answer: C. GM/r^2
• **g = GM/r²** = Gravitational field strength is force per unit mass; from F = GMm/r², dividing by m gives g = GM/r². • **Units: m/s² or N/kg** — Both are equivalent, confirming g is force per unit mass. • 💡 Wrong-option analysis: GM/r: wrong dimension (m²/s² ≠ m/s²); GM²/r²: uses M squared, which has no basis in Newton's law; GMr: wrong dimension and increases with distance, opposite of reality.
The principle of superposition in gravitation states that?
Correct Answer: D. Net gravitational force equals the vector sum of forces due to individual masses
• **Vector sum of all individual forces** = Superposition means each pair of masses interacts independently; the net force on any mass is the vector sum of all pairwise forces. • **Universal applicability** — Essential for calculating the gravitational field of extended bodies like Earth. • 💡 Wrong-option analysis: Depends on shape only: shape affects distribution but superposition applies regardless; Only nearest mass exerts gravity: all masses exert gravity — the Sun still affects us from far away; Depends on temperature: gravitational force has no temperature dependence.
For a spherically symmetric body, the gravitational field outside it behaves as if?
Correct Answer: C. All its mass were concentrated at its center
• **All mass concentrated at its center** = Newton's shell theorem proves that outside a spherically symmetric body, the gravitational field equals that of a point mass equal to the total mass at the center. • **Key result for planets** — This allows Earth (radius 6,400 km) to be treated as a point mass for orbital calculations. • 💡 Wrong-option analysis: Mass spread to infinity: would give a much weaker field; All mass on surface only: shell theorem gives center-concentrated exterior field, not surface-only; Only half mass: no basis for halving the mass in the shell theorem.
The first cosmic velocity for a circular orbit close to Earth’s surface is?
Correct Answer: C. sqrt(gR)
• **v₁ = √(gR)** = For circular orbit just above surface: mg = mv₁²/R → v₁ = √(gR) ≈ 7.9 km/s. • **First cosmic velocity ≈ 7.9 km/s** — The minimum orbital speed for Earth. • 💡 Wrong-option analysis: gR: wrong dimension (m²/s², not m/s); √(2gR): that is escape velocity (second cosmic velocity), √2 times larger; 2πR/g: has dimension of time (seconds), not speed.