mars has a mass 1/10 that of earth and a diameter 1/2 that of earth

Mars has a mass 1/10 that of Earth and a diameter 1/2 that of Earth. The acceleration of a falling body body near the surface of Mars is most nearly:

A. [tex]0.25 m/s^{2}[/tex]

B. [tex]0.5 m/s^{2}[/tex]

C. [tex]2 m/s^{2}[/tex]

D.[tex] 4 m/s^{2}[/tex]

E. [tex]25 m/s^{2}[/tex]

Let's begin by the fact that on Earth, the gravity force [tex]F_{g}[/tex] that acts on a falling object is given by:

[tex]F_{g}=mg_{E}=\frac{Gm_{E}m}{r_{E}^{2}}[/tex] (1)

Where:

[tex]m[/tex] is the mass of the falling body

[tex]m_{E}=5.972(10)^{24} kg[/tex] is the mass of the Earth

[tex]g_{E}[/tex] is the acceleration due gravity on Earth

[tex]G=6.67(10)^{-11} Nm^{2}/kg^{2}[/tex] is the Universal Gravitational constant

[tex]r_{E}=\frac{d_{E}}{2}=6371000m[/tex] is the Earth's radius and [tex]d_{E}=12742000 m[/tex] is its diameter

Simplifying (1) we have:

[tex]g_{E}=\frac{Gm_{E}}{r_{E}^{2}}[/tex] (2)

[tex]g_{E}=\frac{G m_{E}}{(\frac{d_{E}}{2})^{2}}[/tex] (3)

Now, in the case of Mars we have:

[tex]g_{M}=\frac{G m_{M}}{(\frac{d_{M}}{2})^{2}}[/tex] (4)

Where:

[tex]m_{M}=\frac{1}{10}m_{E}[/tex]

[tex]d_{M}=\frac{1}{2}d_{E}[/tex]

Substituting these on (4):

[tex]g_{M}=\frac{G \frac{1}{10}m_{E}}{(\frac{\frac{1}{2}d_{E}}{2})^{2}}[/tex] (5)

Simplifying:

[tex]g_{M}=\frac{8}{5} \frac{G m_{E}}{d_{E}^{2}}[/tex] (6)

Then:

[tex]g_{M}=\frac{8}{5} \frac{(6.67(10)^{-11} Nm^{2}/kg^{2})(5.972(10)^{24} kg)}{(12742000 m)}^{2}}[/tex] (7)

Finally:

[tex]g_{M}=3.92 m/s^{2} \approx 4 m/s^{2}[/tex] This is most nearly the acceleration of a falling body near the surface of Mars