Question

What is the gravitational field strength at the surface of Jupiter (mass $1.9\times {10}^{27}kg$, radius $7.1\times {10}^{7}m$?

• A. $25Nk{g}^{-1}$
• B. $25Nk{g}^{-2}$
• C. $35Nk{g}^{-1}$
• D. $55Nk{g}^{-2}$

Answer 1

The correct option is A $25Nk{g}^{-1}$

Gravitational field = $K\frac{m1}{r^2}$

where $K$ is universal gravitational constant $m_1$ is the mass $r$ is the distance.

G= $6.67\times {10}^{-11}\frac{1.9\left({10}^{27}\right)}{{(7.1\times {10}^7)}^2}$

= $25N{kg}^{-1}$