Question

What is the free-fall acceleration at the surface of (a) the moon and (b) Jupiter?

Short answer

The free-fall acceleration at the surface of (a) the moon is $1.62\mathrm{m}/{\mathrm{s}}^2$; and (b) Jupiter is$24.8\mathrm{m}/{\mathrm{s}}^2.$

Step-by-step solution

Given information (a).

Universal gravitational constant: $\mathrm{G}=6.67\times {10}^{-11}\mathrm{Nm}2/\mathrm{kg}2.$

Mass of moon: $Mm=7.34\times {10}^{24}\mathrm{kg}.$

Radius of moon $\mathrm{rm}=1.74\times {10}^6\mathrm{m}.$

Calculation (a).

The formula of free-fall acceleration is given by : $g=GMR2.$

The free-fall acceleration at the on the surface of moon is :

localid="1648481120225" $$\begin{gathered} gm=GMmrm2 \\ =(6.67\times {10}^{-11}\mathrm{Nm}2/\mathrm{kg}2)(7.34\times {10}^{24}\mathrm{kg})(1.74\times {10}^6\mathrm{m})2. \\ =1.62\mathrm{m}/{\mathrm{s}}^2. \end{gathered}$$

Final answer (a).

The free-fall acceleration at the on the surface of moon is$1.62m/s^2.$

Given information (b).

Universal gravitational constant: $\mathrm{G}=6.67\times {10}^{-11}\mathrm{Nm}2/\mathrm{kg}2.$

Mass of Jupiter: $Mj=1.90\times {10}^{27}\mathrm{kg}.$

Radius of Jupiter: $rj=6.99\times {10}^7\mathrm{m}.$

Calculation (b).

The free-fall acceleration at the surface of Jupiter is :

$$\begin{gathered} gj=GMjrj2 \\ =(6.67\times {10}^{-11}N{m}^2/k{g}^2)(1.9\times {10}^{27}kg)(6.99\times {10}^{7}m)2 \\ =24.8m/s^2. \end{gathered}$$

Final answer (b).

The free-fall acceleration at the surface of Jupiter is$24.8\mathrm{m}/{\mathrm{s}}^2.$