3 years ago

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

Physics

1 answer:

Rufina [12.5K]3 years ago

4 0

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:

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

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

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

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

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3 years ago

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kifflom [539]

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4 years ago

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4 years ago

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nikitadnepr [17]

The answer is C , you’re welcome

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