Gravitation — Set 2

Correct Answer: B. sqrt(GM/r)

• **v = √(GM/r)** = Setting gravity equal to centripetal force: GMm/r² = mv²/r → v = √(GM/r). • **Decreases with orbital radius** — The ISS at ~400 km altitude orbits at ≈ 7.7 km/s; higher satellites move slower. • 💡 Wrong-option analysis: √(2GM/r): that is escape velocity, √2 times the orbital speed; √(GMr): wrong dimension — gives m²/s not m/s; √(GM)·r: dimensionally wrong and increases with r, opposite of reality.

Correct Answer: B. 2π sqrt(r^3/GM)

• **T = 2π√(r³/GM)** = From T = 2πr/v with v = √(GM/r); this is Kepler's third law in exact form. • **T²/r³ = 4π²/GM** — The ratio is constant for all satellites orbiting the same planet. • 💡 Wrong-option analysis: 2π√(GM/r): wrong units, not a time period; 2π√(GMr³): dimensionally incorrect; 2π√(GMr): does not give correct time units.

Correct Answer: D. They are in continuous free fall with the spacecraft

• **Continuous free fall** = Both astronaut and spacecraft fall freely under gravity together; no normal reaction force means apparent weightlessness. • **g ≈ 8.7 m/s² at ISS altitude (400 km)** — Gravity is still strong there; the ISS is simply falling around Earth. • 💡 Wrong-option analysis: No gravity in space: g ≈ 89% of surface value at 400 km; Air resistance cancels gravity: there is essentially no air at orbital altitude; Mass becomes zero: mass never changes — only apparent weight is zero.

Correct Answer: D. g decreases approximately as (1 - d/R)

• **g′ ≈ g(1 − d/R)** = Inside a uniform sphere, only the mass within radius (R−d) contributes; as depth d increases, enclosed mass decreases, so g decreases linearly. • **At Earth's center, g = 0** — All gravitational pulls cancel by symmetry at the exact center. • 💡 Wrong-option analysis: g stays constant: gravity decreases going underground; g increases as (1+d/R): contradicts the shell theorem; g becomes infinite at center: the opposite is true — g → 0 at center.

Correct Answer: B. -GMm/r

• **U = −GMm/r** = U = mV = m(−GM/r); the negative sign indicates a bound, attractive system. • **Increases toward zero at infinity** — As r → ∞, U → 0; closer objects have more negative (lower) potential energy. • 💡 Wrong-option analysis: +GMm·r: positive and increasing with distance — wrong for attraction; −GMm·r: wrong dimension (J·m instead of J); +GMm/r: positive value would imply repulsion, not attraction.

Correct Answer: C. One-fourth

• **One-fourth** = Since F ∝ 1/r², doubling r gives F → F/(2²) = F/4. • **Inverse-square law** — This is the defining property of both gravitational and electrostatic forces. • 💡 Wrong-option analysis: Half: would apply only if F ∝ 1/r, not 1/r²; Double: doubling distance never increases an attractive force; Four times: force increases four-fold only when distance is halved, not doubled.

Correct Answer: D. Independent of its mass

• **Independent of mass** = In free fall, force = mg and mass = m, so acceleration = mg/m = g; mass cancels completely. • **Galileo's result (≈ 1590)** — All objects fall at the same rate when air resistance is removed. • 💡 Wrong-option analysis: Zero for large mass: all masses fall with g ≈ 9.8 m/s² near Earth; Greater for lighter bodies: less force (mg) balances less inertia (m) exactly; Greater for heavier bodies: same reasoning — extra force exactly matches extra inertia.

Correct Answer: D. Zero

• **Zero at center** = At the exact center of a uniform sphere, gravitational pulls from all directions are equal and opposite, canceling by symmetry. • **g varies linearly with r inside** — From 0 at center to GM/R² at the surface, for a uniform solid sphere. • 💡 Wrong-option analysis: Infinite: requires all mass at a point, which is not the case; Equal to g at surface: surface value is GM/R², center value is zero; Maximum: g is maximum at the surface, not the center, for a uniform solid sphere.

Correct Answer: D. It is negative

• **Negative gravitational PE** = Taking U = 0 at infinity, U = −GMm/r is always negative for finite r; the satellite is bound to the planet. • **Total energy also negative** — E = −GMm/(2r) < 0 confirms a bound orbit; E = 0 means just barely escaping. • 💡 Wrong-option analysis: Zero for any orbit: U = 0 only at infinite separation; Always positive: would need a different reference convention; Depends only on satellite shape: shape does not appear in U = −GMm/r.

Correct Answer: A. Its period is 24 hours and it orbits in the equatorial plane

• **24-hour period in equatorial plane** = A geostationary satellite must match Earth's rotation (≈24 h) and orbit in the equatorial plane to remain fixed over one ground point. • **Altitude ≈ 35,786 km** — The only orbital radius where period equals Earth's rotation period. • 💡 Wrong-option analysis: Any tilted plane: tilted orbit traces a figure-eight over Earth, not a fixed point; Orbits over poles: polar orbits cannot be geostationary; Period less than 12 hours: lower orbit means faster speed — satellite would not stay above one point.