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!    

Answer 2

The acceleration due to the gravity on the surface of Mars is most nearly equal to 2/5g. hence, option (c) is correct.

Given data:

The ratio of Earth's radius and Mars radius is, r/r' = 19/10.

The ratio of density of Earth and Mars is, $\rho/\rho'=14/10$.

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

For Earth,

$\rho = \dfrac{m}{V}= \dfrac{m}{\dfrac{4}{3}\pi r^{3}}$ ....................................................................(1)

And for Mars,

$\rho' = \dfrac{m'}{V'}\\\\\rho' = \dfrac{m'}{\dfrac{4}{3}\pi r'^{3}}$ .............................................................................(2)

Then take the ratio of equation (1) and (2) as,

$\dfrac{\rho}{\rho'}=\dfrac{m \times r^{3}}{m' \times r'^{3}}= \dfrac{14}{10}$

Now substituting the values as,

$\dfrac{m}{m'}=\dfrac{r^{3}}{ r'^{3}} \times \dfrac{14}{10}\\\\\dfrac{m}{m'}=\dfrac{19^{3}}{ 10^{3}} \times \dfrac{14}{10}\\\\\dfrac{m}{m'}=\dfrac{19^{3}}{ 10^{3}} \times \dfrac{14}{10}$ .....................................................................(3)

Now, apply the formula for the gravitational acceleration for Earth and Mars as,

$g= \dfrac{Gm}{r^{2}}\\\\g'= \dfrac{Gm'}{r'^{2}}$

taking the ratio and substitute the value of equation (3) as,

$\dfrac{g}{g'}= \dfrac{\dfrac{Gm}{r^{2}}}{\dfrac{Gm'}{r'^{2}}}\\\\\\\dfrac{g}{g'}= \dfrac{m \times r'^{2}}{m' \times r^{2}}\\\\\\\dfrac{g}{g'}= \dfrac{m \times r'^{2}}{m' \times r^{2}}\\\\\dfrac{g}{g'}=\dfrac{19^{3}}{ 10^{3}} \times \dfrac{14}{10} \times \dfrac{10^{2}}{19^{2}}\\\\\\g' = 0.38g\\\\g' \approx 2/5g$

Thus, we can conclude that the acceleration due to the gravity on the surface of Mars is most nearly equal to 2/5g. hence, option (c) is correct.

Learn more about the gravitational acceleration here:

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