Question

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

Answer 1

24.79 m/s2

Step-by-step 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(mM_j)/(r_j^(2) )`

`g_j=G(M_j)/(r_j^(2) )`

`g_j=6.67*10^(-11) (1.9*10^(27) )/(71492000^(2) )`

`g_j=24.79m/s^(2)`