Question

Suppose all the mass of the Earth were compacted into a small spherical ball.

What radius must the sphere have so that the acceleration due to gravity at the Earth's new surface was equal to the acceleration due to gravity at the surface of the Sun?
( MSun=1.99×1030kg , RSun=6.96×108m )

Express your answer with the appropriate units.

Answer 1

F = G M m / R^2 a = G M / R^2

aS / aE = (MS / RS^2) / (ME / RE^2)

aS / aE = 329390 * RE^2 / RS^2) = 1 If acceleration due gravity equal

RE^2 = 1 / 329290 * (6.96E8)^2

RE^2 = 1.471 E12

RE = 1.213E6 m

actual RE = 6.38E6 m (average density of sun < that of earth)