Question

A 7.8 ×× 102121 kgkg moon orbits a distant planet in a circular orbit of radius 1.5 ×× 1088 mm. It experiences a 1.1 ×× 101919 NN gravitational pull from the planet.

Part A What is the moon's orbital period in earth days?

Answer 1

In a stable orbit, the gravitational pull due to the planet on the moon will be equal and opposite to the centrpetal force it experiences due to its orbital velocity and orbit length i.e

= 1.1 x 1019 N GMm mv2 R2 R Given GMm The gravitational force of the planet i.e R2 tertThemassof themoon : m =7.8 x 1021 Kg The orbital radius: R=1.5 x 108 m Thus 7.8 x 1021 x v2 1.1 x 1019 1.5 x 108 1.1 x 1019 x 1.5 x 108 v2 7.8 x 1021

$\\ v=\sqrt{\frac{1.1\times 10^{19}\times 1.5\times 10^8}{7.8\times 10^{21}}} \\ v=459.93 \ \text{m/s} \\ \text{Now} \ T=\frac{2\pi R}{v} \\ T=\frac{2\times 3.142 \times 1.5\times 10^8}{459.93} \\ \text{To find the above time in earth days we have to find}\ \frac{T}{24\times 60 \times 60} \\ \text{Thus }\ T_{ans}=\frac{2\times 3.142 \times 1.5\times 10^8}{459.93\times 24\times 60 \times 60 } \\ \text{Answer} \ T_{ans}=23.72 \ \text{Earth days}$