Gravitation — Set 4
Physics · गुरुत्वाकर्षण · Questions 31–40 of 60
The ratio of escape velocity to the circular orbital velocity near a planet’s surface is?
Correct Answer: C. sqrt(2)
• **Ratio = √2** = v_escape = √(2GM/R) and v_orbit = √(GM/R); their ratio = √(2GM/R)/√(GM/R) = √2. • **Practical note** — To escape Earth from orbit, a spacecraft needs only about 41% more speed. • 💡 Wrong-option analysis: 1/2: inverts the ratio — escape speed is greater than orbital speed; 2: ratio of the formulas is √2, not 2; 1: equal only if the factor of 2 inside the square root is ignored.
If the mass of a planet is doubled while its radius remains the same, its escape velocity becomes?
Correct Answer: A. sqrt(2) times
• **√2 times** = v_e = √(2GM/R); M → 2M gives v_new = √(2G·2M/R) = √2 × v_original. • **Square-root dependence on M** — A 4× increase in mass doubles escape velocity. • 💡 Wrong-option analysis: Half: escape velocity never decreases when mass increases; Double: would require M to increase 4-fold (since v ∝ √M); Unchanged: escape velocity clearly depends on M and changes.
A 60 kg person stands on the Moon where g ≈ 1.62 m/s^2. What is the person’s weight there (approximately)?
Correct Answer: B. 97 N
• **W ≈ 97 N** = W = mg = 60 × 1.62 = 97.2 N; mass stays 60 kg but weight changes with g. • **Moon's g ≈ 1/6 of Earth's** — A person weighing 588 N on Earth weighs only ≈ 97 N on the Moon. • 💡 Wrong-option analysis: 60 N: confuses kg (mass) with N (weight); 588 N: that is weight on Earth (60 × 9.8), not the Moon; 16.2 N: incorrectly uses 1.62 × 10 instead of 60 × 1.62.
A 1 kg mass at Earth’s surface has gravitational force roughly equal to?
Correct Answer: A. 9.8 N
• **9.8 N** = W = mg = 1 kg × 9.8 m/s² = 9.8 N; approximately 1 kilogram-force. • **g ≈ 9.80665 m/s²** — Standard SI value of acceleration due to gravity at Earth's surface. • 💡 Wrong-option analysis: 98 N: weight of a 10 kg mass; 0.98 N: ten times too small; 980 N: weight of a 100 kg mass.
Gravitational potential at a point is defined as?
Correct Answer: C. Work done per unit mass in bringing a test mass from infinity to that point
• **Work per unit mass from infinity** = V = W/m where W is work done by gravity bringing unit mass from infinity; V = −GM/r. • **Units: J/kg** — Potential is negative, meaning gravity does positive work as the mass moves inward. • 💡 Wrong-option analysis: Work done moving to infinity: that gives negative of potential, not potential itself; Force per unit mass: that is gravitational field strength g, not potential V; Energy per unit volume: energy density (J/m³), not J/kg.
The value of the universal gravitational constant G in SI units is approximately?
Correct Answer: B. 6.67×10^-11
• **G ≈ 6.67×10⁻¹¹ N m² kg⁻²** = First measured by Henry Cavendish in 1798 using a torsion balance; one of the fundamental constants of nature. • **SI unit: N m² kg⁻²** — G appears in F = Gm₁m₂/r². • 💡 Wrong-option analysis: 6.67×10⁻⁹: two orders of magnitude too large; 9.8: that is g (acceleration), not G (the constant); 3.00×10⁸: the speed of light c, not the gravitational constant.
A geostationary satellite appears fixed over Earth because its orbital period is?
Correct Answer: D. Equal to Earth’s rotational period (about 24 hours)
• **24-hour orbital period** = A geostationary satellite must complete one orbit in the same time Earth completes one rotation (≈23 h 56 min ≈ 24 h). • **Altitude ≈ 35,786 km** — Communications and weather satellites use geostationary orbits. • 💡 Wrong-option analysis: 6 hours: too fast — satellite would sweep ahead of Earth's rotation; 30 days: that is the Moon's period, not geostationary; 12 hours: completes two revolutions per day, appearing to oscillate.
For a uniform solid sphere of density ρ and radius R, the surface gravity g is?
Correct Answer: C. (4/3)πGρR
• **g = (4/3)πGρR** = Substituting M = (4/3)πR³ρ into g = GM/R² gives g = (4/3)πGρR. • **Proportional to R for same density** — A planet twice the radius (same density) has twice the surface gravity. • 💡 Wrong-option analysis: Gρ/R: lacks the factor (4/3)π needed to relate density to mass; GM/R²: correct in terms of M but the density form was requested; GM/R: wrong dimension (m²/s² ≠ m/s²).
If the acceleration due to gravity g decreases, the time period of a simple pendulum (same length) will?
Correct Answer: D. Increase
• **Period increases** = T = 2π√(L/g); if g decreases, √(L/g) becomes larger, so T increases. • **On Moon (g = 1.62 m/s²)** — Same-length pendulum period is about 2.46 times longer than on Earth. • 💡 Wrong-option analysis: Remain same: T depends on g; Become zero: requires infinite g; Decrease: smaller g makes √(L/g) larger, not smaller.
The total mechanical energy E of a satellite of mass m in a circular orbit of radius r around mass M is?
Correct Answer: B. −GMm/(2r)
• **E = −GMm/(2r)** = KE = GMm/(2r) and PE = −GMm/r; total E = GMm/(2r) − GMm/r = −GMm/(2r). • **Negative total energy** — Confirms the satellite is bound; E = 0 corresponds to escape. • 💡 Wrong-option analysis: −GMm/r: that is only the potential energy; 0: zero total energy = parabolic escape trajectory; +GMm/r: positive energy means the satellite can escape.