Question

The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R, the radius of the planet would be(a) $\frac{1}{2}R$ (b) 2R(c) 4R(d) $\frac{1}{4}R$

Answer 1

Answer:

( a ) ( 1 / 2 ) R

Explanation:

Let the mass of earth be M1 and that of the planet be M2.

Let the radius of the planet be R2.

Volume = ( 4 / 3 ) × π × r^3

Since density of the planet is twice that of earth,

2 [ M1 / { ( 4 / 3 ) × π × R ^3 } ] = [ M2 / { ( 4 / 3 ) × π × ( R2 ) ^3 ]

2 [ M1 / R ^3 ] = M2 / ( R2 ) ^3

2 M1 / R^3 = M2 / ( R2 ) ^3

Now, mass = ( g r ^2 ) / G

( As g = ( G M ) / r ^2 )

Since g on both planets is equal,

[ ( 2 g R ^2 ) / G ] / R ^3 = [ ( g R2 ^2 ) / G ] / R2 ^3

[ ( 2 R^2 ) ] / R ^3 = [ R2 ^2 ] / R2 ^3

2 / R = 1 / R2

R2 = R / 2

Therefore, answer = ( 1 / 2 ) R

Answer 2

Explanation:

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