# A orbiting satellite stays over a certain spot on the equator of (rotating) Neptune. What is the...

## Question:

A orbiting satellite stays over a certain spot on the equator of (rotating) Neptune. What is the altitude of the orbit (called a "synchronous orbit")? (The radius of Neptune is 24750 km.)

## Kepler's Third Law:

The square of time period of revolution of a planet round the sun is directly proportional to the cube of the mean distance between the earth and sun. This law is also valid for a a satellite revolving round a planet.The constant of proportionality depends on the mass of planet. The altitude of the satellite is the difference of orbital radius and the radius of the planet.

Given:

Radius of Neptune given, {eq}R = 24750 km= 2.4 \times 10^7\ m {/eq}

Time taken by Neptune to complete one rotation, {eq}T = 16\ Hrs = 16 \times 3600\ s = 5.76 \times 10^4\ s {/eq}

Mass of Neptune, {eq}M = 102.4 \times 10^{24}\ kg {/eq}

Let the altitude be H and is given by, {eq}H = r + R {/eq}

where R is radius of Neptune and r is orbital radius given by

{eq}r^3 = \dfrac{G M }{4 \pi^2 } T^2 \\ r^3= \dfrac{6.67 \times 10^{-11} \times 102.4 \times 10^{24}}{ 4 \times \pi^2 } (5.76 \times 10^4 s)^2 \\ r^3 = 5.746 \times 10^{24} \\ r= 1.79 \times 10^8 \\ H = 1.79 \times 10^8 - 2.4 \times 10^7\ m \\ H = 20.3 \times 10^7\ m {/eq}