Question

A planet has twice the density of earth but the acceleration due to gravity on its surface is exactly the same as that on the surface of earth. Its radius in terms of earth R will be

• A. R/4
• B. R/2
• C. R/3
• D. R/8.

Answer 1

Answer: (B)

R/2

Answer 2

We know that

g = $\frac { \mathrm { GM } } { \mathrm { R } ^ { 2 } }$ = $\frac { \mathrm { G } } { \mathrm { R } ^ { 2 } } \left[ \frac { 4 } { 3 } \pi \mathrm { R } ^ { 3 } \mathrm {~d} \right]$

where d = mean density of earth.

For planet

g′ = $\frac { \mathrm { G } } { \left( \mathrm { R } ^ { \prime } \right) ^ { 2 } } \times \left[ \frac { 4 } { 3 } \pi \mathrm { R } ^ { \prime 3 } ( \mathrm {~d} ) \right]$

Given that g = g′

Therefore,

=$\frac { \mathrm { G } } { \mathrm { R } ^ { 2 } } \left[ \frac { 4 } { 3 } \pi \mathrm { R } ^ { 3 } \mathrm {~d} \right]$ = $\frac { \mathrm { G } } { \left( \mathrm { R } ^ { \prime } \right) ^ { 2 } } \times \frac { 4 } { 3 } \pi \mathrm { R } ^ { 3 } ( 2 \mathrm {~d} )$

Solving we get, R′ = (R/2).

Therefore (2)