Question

What is the free-fall acceleration at the surface of the jupiter?

Answer 1

Answer:

24.79 m/s2

Explanation:

Let us assume that there is an object with a mass of 'm' on the

 

Jupiter. Jupiter will attract this object:

$mg_j=G\frac{mM_j}{r_j^{2} }$

$g_j=G\frac{M_j}{r_j^{2} }$

$g_j=6.67*10^{-11} \frac{1.9*10^{27} }{71492000^{2} }$

$g_j=24.79m/s^{2}$