The Martian moon Phobos travels in a circular orbit of radius $9.4 imes 10^6$ m around Mars with a period of 7.6 hours - Leaving Cert Physics - Question b - 2014

The formula derived from Kepler's Third Law is:

$T^2 = \frac{4\pi^2 R^3}{G M}$

Where:

• $T$ is the orbital period (in seconds),
• $R$ is the radius (in meters),
• $G$ is the gravitational constant $6.67 \times 10^{-11} \text{ N(m/kg)}^2$, and
• $M$ is the mass of Mars.

Rearranging the formula gives:

$M = \frac{4\pi^2 R^3}{G T^2}$

Substituting the values:

• $R = 9.4 \times 10^6$ m
• $T = 27360$ s

We find:

$M = \frac{4\pi^2 (9.4 \times 10^6)^3}{(6.67 \times 10^{-11})(27360^2)}$

Now, performing the calculations:

1. Calculate $R^3$:

   • $(9.4 \times 10^6)^3 \approx 8.285 \times 10^{20}$ m³

2. Calculate $T^2$:

   • $27360^2 \approx 7.469 \times 10^8$ s²

3. Substitute in the values:

$M = \frac{4\pi^2 (8.285 \times 10^{20})}{(6.67 \times 10^{-11})(7.469 \times 10^8)}$

Calculating this gives:

$M \approx 6.57 \times 10^{23} ext{ kg}$