Question

The ratio of Earth's radius to that of Mars (RR/RM) is approximately 19/10, and the ratio of the density of Earth to that of Mars (PE/PM) is approximately 14/10. Assume both planets are solid spheres of uniform density. The acceleration due to gravity on Earth is g. The acceleration due to the gravityon the surface of Mars is most nearly:______.
a. 1/5 g.
b. 1/4 g.
c. 2/5 g.
d. 4/3 g.
e. 2 g.

Answer 1

Answer:

The answer is letter c.

Explanation:

We need to find the relation of masses first and then apply the equivalence principle.

We know that the ratio of the density of Earth to that of Mars is approximately 14/10.

$\frac{\rho_{E}}{\rho_{M}}=\frac{14}{10}$

Density is the mass divided by the volume (Here the volume of each planet is a solid sphere of uniform density), so

$\rho_{E}=\frac{M_{E}}{V_{E}}=\frac{M_{E}}{\frac{4}{3}\pi R_{E}^{3}}$

$\rho_{M}=\frac{M_{M}}{V_{M}}=\frac{M_{E}}{\frac{4}{3}\pi R_{M}^{3}}$

Then, the ratio between the densities will be:

$\frac{\rho_{E}}{\rho_{M}}=\frac{M_{E}R_{M}^{3}}{M_{M}R_{E}^{3}}=\frac{14}{10}$

We know that $\frac{R_{E}}{R_{M}}=19/10$, then we can write the above equation as:

$\frac{M_{E}}{M_{M}}=\left(\frac{R_{E}}{R_{M}}\right)^{3} \frac{14}{10}$

$\frac{M_{E}}{M_{M}}=(\frac{19}{10})^{3} \frac{14}{10}$

Now, using the gravitational force:

To the earth                                

$g=G\frac{M_{E}}{R_{E}^{2}}$

To the mars

$g=G\frac{M_{M}}{R_{M}^{2}}$

But, we know that $M_{M}=(\frac{10}{19})^{3} \frac{10}{14}M_{E} \: and \: R_{M}=\frac{10}{19}R_{E}$

Therefore, we will have:

$g=G\frac{(\frac{10}{19})^{3} \frac{10}{14}M_{E}}{(\frac{10}{19}R_{E})^{2}}$

$g=\frac{10*10}{14*19}G\frac{M_{E}}{R_{E}^{2}}$      

$g_{M}=0.38g$

$g_{M} \approx \frac{2}{5}g$    

The answer is letter c.

I hope it helps you!