Q.52 What is the acceleration due to gravity (m/s2) on the surface of a planet if its
radius is 1/4th that of earth and its mass is 1/80th that of earth? Assume that the
gravity on the surface of the earth is 10 m/s2.
The acceleration due to gravity on this planet is 2 m/s². This result follows directly from Newton’s law of universal gravitation applied to planetary surfaces. Given Earth’s gravity as 10 m/s², the calculation accounts for the planet’s reduced mass and radius.
Core Formula
Acceleration due to gravity g on a planet’s surface uses g = GM/R², where G is the gravitational constant, M is mass, and R is radius.
For Earth, ge = GMe/Re² = 10 m/s². For the planet, gp = GMp/Rp², with Mp = Me/80 and Rp = Re/4.
The ratio simplifies to gp/ge = (Mp/Me) × (Re/Rp)².
Step-by-Step Calculation
Substitute ratios: gp/ge = (1/80) × (1/(1/4))² = (1/80) × (4)² = (1/80) × 16 = 16/80 = 0.2.
Thus, gp = 0.2 × 10 = 2 m/s². Smaller mass decreases gravity, but smaller radius (squared in denominator) increases it; net effect is 1/5th of Earth’s.
ge = 10; Rp = 1/4; Mp = 1/80; gp = ge * Mp / (Rp**2) → 2.0
Option Analysis (MCQ Context)
Typical CSIR NET choices might include 0.5, 2, 5, 8, 20 m/s². 2 m/s² is correct as mass factor (1/80) dominates radius factor (16), yielding 16/80=0.2 times ge.
[memory:4]
- 0.5 m/s²: Common error forgetting R² (using linear radius: 1/80 × 4 = 0.05 → 0.5 m/s²).
- 5 m/s²: Inverting ratios (80/16=5 → 50 m/s², or similar miscalculation).
- 2 m/s²: Correct; proper ratio method.
Always square radius term; mass scales linearly.
CSIR NET Exam Tips
This tests gravitation ratio method, avoiding G value recall. Memorize g ∝ M/R² for quick solves.
Similar problems compare Moon (R≈1/4 Re, M≈1/80 Me, gmoon≈ge/5).
Practice variations: if radius halves and mass quarters, g doubles (1/4 × 4 = 1 → same g).
