Question

The planet Mars has two moons, if one of them has a period 7 hours, 30 minutes and an orbital radius of $9.0 \times 10^3$ km. Find the mass of Mars.
$\left\{ \text{Given } \frac{4\pi^2}{G} = 6 \times 10^{11} \text{ N}^{-1} \text{ m}^{-2} \text{ kg}^2 \right\}$

• A. $5.96 \times 10^{19}$ kg
• B. $3.25 \times 10^{21}$ kg
• C. $6.00 \times 10^{23}$ kg
• D. $7.02 \times 10^{25}$ kg

Hint:

Always convert all given quantities to their base SI units (meters, kilograms, seconds) before substituting them into a physics formula. The constant being provided as a group ($\frac{4\pi^2}{G}$) is a common shortcut in exams to simplify calculations.

Answer 1

The Correct Option is C

For a moon orbiting a planet, the gravitational force provides the necessary centripetal force.
$F_g = F_c$
$\frac{GMm}{r^2} = m\omega^2r$
Where M is the mass of Mars, m is the mass of the moon, r is the orbital radius, and $\omega$ is the angular velocity.
$GM = \omega^2r^3$.
We know that angular velocity $\omega = \frac{2\pi}{T}$, where T is the time period.
$GM = \left(\frac{2\pi}{T}\right)^2 r^3 = \frac{4\pi^2r^3}{T^2}$.
Rearranging the formula to solve for the mass of Mars (M):
$M = \frac{4\pi^2 r^3}{G T^2} = \left(\frac{4\pi^2}{G}\right) \frac{r^3}{T^2}$.
First, convert the given quantities to SI units.
Orbital radius $r = 9.0 \times 10^3 \text{ km} = 9.0 \times 10^6 \text{ m}$.
Time period T = 7 hours 30 minutes.
$T = (7 \times 3600) + (30 \times 60) = 25200 + 1800 = 27000 \text{ s} = 2.7 \times 10^4 \text{ s}$.
Now, substitute the values into the equation for M.
$M = (6 \times 10^{11}) \times \frac{(9.0 \times 10^6)^3}{(2.7 \times 10^4)^2}$.
$M = (6 \times 10^{11}) \times \frac{729 \times 10^{18}}{7.29 \times 10^8}$.
$M = (6 \times 10^{11}) \times \left(\frac{729}{7.29}\right) \times \frac{10^{18}}{10^8}$.
$M = (6 \times 10^{11}) \times 100 \times 10^{10}$.
$M = 6 \times 10^{11} \times 10^2 \times 10^{10} = 6 \times 10^{23} \text{ kg}$.