Question

At what fraction of its current radius would the free-fall acceleration at the surface be three times its present value?

Answer 1

Answer:

0.577

Explanation:

The initial weight is

mg= $\frac{GMm}{r_1^2}$.......................1

and the final weight is

$3mg= \frac{GMm}{r_2^2}$......................2

now to calculate $r_2/r_1$

now dividing equation 1 by equation 2 we get

$\frac{1}{r_1^2}/\frac{1}{r_2^2}$= 1/3

$\frac{r_2}{r_1} =\sqrt{\frac{1}{3} }$

$\frac{r_2}{r_1}$=0.577