Question

The moon moves around the earth in a circular orbit with a period of 27 days. If $R = 6.4 \times 10^6$ m is Earth's radius and $g = 9.8$ m/s$^2$, and if $D$ is the moon's orbital radius, find $D/R$. (Specify answer up to one digit after decimal point.)

Hint:

Use $GM = gR^2$ to avoid calculating Earth's mass explicitly.

Answer 1

Correct Answer: 59

Step 1: Use centripetal force balance.
$\frac{GM}{D^2} = \frac{4\pi^2 D}{T^2}$.

Step 2: Replace $GM$ using surface gravity.
At Earth's surface: $g = \frac{GM}{R^2} $\Rightarrow$ GM = gR^2$.

Step 3: Substitute into orbital equation.
$\frac{gR^2}{D^2} = \frac{4\pi^2 D}{T^2}$.
Solve: $D^3 = \frac{gR^2 T^2}{4\pi^2}$.

Step 4: Convert period to seconds.
$T = 27$ days = $27 \times 24 \times 3600 = 2.3328\times 10^6$ s.

Step 5: Compute $D$.
$D^3 = \frac{9.8 (6.4\times10^6)^2 (2.3328\times10^6)^2}{4\pi^2}$.
$D \approx 3.86\times10^8$ m.

Step 6: Compute ratio.
$\frac{D}{R} = \frac{3.86\times10^8}{6.4\times10^6} = 60.3$.