Gravitation — Set 1

Correct Answer: B. Product of the two masses and inversely proportional to the square of the distance

• **Product of the two masses (m₁×m₂)** = Newton's law states F = Gm₁m₂/r², so force is proportional to the product of both masses. • **Inverse-square law (1/r²)** — The exponent 2 distinguishes gravity from other force laws and explains planetary orbits. • 💡 Wrong-option analysis: Sum of masses: force would not triple when one mass triples; Difference of masses: would give zero force for equal masses, which is wrong; Inversely proportional to distance (not squared): incompatible with Kepler's third law.

Correct Answer: A. [M^-1 L^3 T^-2]

• **[M⁻¹ L³ T⁻²]** = From G = Fr²/(m₁m₂), substituting [MLT⁻²] for force and [L²] for r² over [M²] gives [M⁻¹L³T⁻²]. • **G = 6.674×10⁻¹¹ N m² kg⁻²** — Its tiny value makes gravitational force negligible between lab-scale objects. • 💡 Wrong-option analysis: [MLT⁻²]: dimension of force, not G; [M⁻¹L²T⁻²]: wrong power of L — one factor of length missing; [M¹L²T⁻¹]: positive mass power contradicts G = Fr²/(m₁m₂) requiring M⁻¹.

Correct Answer: A. Weight equals mg and changes if g changes

• **Weight = mg** = Weight is the gravitational force W = mg; it changes whenever g changes with location. • **Mass (kg) is invariant** — Mass measures amount of matter and stays the same on Earth, Moon, or in space. • 💡 Wrong-option analysis: Mass equals mg: mass is not a force product; Weight is constant everywhere: weight decreases at altitude and on other planets; Mass equals weight in newtons: mass is in kg, weight in N — they are different quantities.

Correct Answer: B. sqrt(2GM/R)

• **v_escape = √(2GM/R)** = Setting kinetic energy ½mv² equal to |gravitational PE| = GMm/R and solving gives √(2GM/R). • **Factor of 2** — This distinguishes escape velocity from orbital speed √(GM/R); Earth's escape velocity ≈ 11.2 km/s. • 💡 Wrong-option analysis: √(GM/R): this is orbital (first cosmic) speed, not escape speed; √(GMR): dimensionally incorrect, gives m²/s not m/s; √(GM/2R): smaller than orbital speed — physically impossible for escape.

Correct Answer: B. g

• **g′ ≈ g(1 − 2h/R)** = From g′ = gR²/(R+h)², using the binomial approximation for h << R gives g′ ≈ g(1 − 2h/R). • **Decreases with altitude** — Each 1 km rise near Earth's surface reduces g by about 0.003 m/s². • 💡 Wrong-option analysis: g(1 + 2h/R): wrong sign — g decreases with height, not increases; g(R/R+h): first-order gives 1−h/R, not 1−2h/R; g(R+h)/R: exceeds g, contradicting the inverse-square law.

Correct Answer: B. -GM/r

• **V = −GM/r** = Gravitational potential is work done per unit mass bringing a test mass from infinity; it is always negative for attractive gravity. • **Negative sign** — Reflects that gravity is attractive and bound systems have lower energy than infinity. • 💡 Wrong-option analysis: −GMr: wrong units (J·m/kg instead of J/kg); +GM/r: positive potential would imply repulsive gravity; +GMr: both wrong sign and wrong dimension.

Correct Answer: A. It is zero everywhere inside

• **Zero everywhere inside** = By the shell theorem, gravitational contributions from all parts of a uniform spherical shell cancel exactly at every interior point. • **Shell theorem (Newton, 1687)** — Earth's hollow regions exert no net gravity on objects inside them. • 💡 Wrong-option analysis: Maximum at center: g decreases toward center inside a solid sphere, not a hollow one; GM/R² everywhere inside: that is the surface value, valid only outside; Varies linearly with distance: linear variation applies inside a solid uniform sphere, not a hollow shell.

Correct Answer: B. g = -dV/dr

• **g = −dV/dr** = Gravitational field is the negative gradient of potential; the negative sign ensures the field points toward decreasing potential (toward the mass). • **Gradient relation** — Analogous to E = −dV/dr in electrostatics. • 💡 Wrong-option analysis: g = V/r: not the general relation, only coincidentally correct for a point mass in specific forms; g = Vr: wrong dimensions; g = +dV/dr: positive gradient would point away from mass, contradicting attraction.

Correct Answer: A. Zero

• **Zero** = For a conservative force, work depends only on start and end positions; over a closed path the net work is zero. • **Path-independence** — This property allows gravitational potential energy to be defined unambiguously. • 💡 Wrong-option analysis: Negative: work can be negative on part of the path but not the full closed loop; Positive: same reasoning — net is zero; Dependent on path length: that describes non-conservative forces like friction.

Correct Answer: C. T^2 is proportional to r^3 for planetary orbits

• **T² ∝ r³** = Kepler's third law: the square of the orbital period is proportional to the cube of the semi-major axis for orbits around the same central mass. • **T²/r³ = 4π²/GM** — This constant ratio allows the Sun's mass to be determined from planetary data. • 💡 Wrong-option analysis: Planets move in circles: Kepler's first law says ellipses; Speed is constant: Kepler's second law says speed varies along the orbit; Gravitational force proportional to distance: Newton's law gives F ∝ 1/r², not r.