Question

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

Answer 1

0.577

Step-by-step explanation:

The initial weight is

mg=
`(GMm)/(r_1^2)`.......................1

and the final weight is

`3mg= (GMm)/(r_2^2)`......................2

now to calculate
`r_2/r_1`

now dividing equation 1 by equation 2 we get

`(1)/(r_1^2)/(1)/(r_2^2)`= 1/3

`(r_2)/(r_1) =\sqrt{(1)/(3) }`

`(r_2)/(r_1)`=0.577