Question

Calculate the height from the Earth's surface that a satellite must attain in order to be in geosynchronous orbit and the satellite's velocity?

Answer 1

Step 1: Given data

The radius of the Earth $\left(\mathrm{R}\right)=6.38\times 106\mathrm{m}$
Gravitation Constant $\left(\mathrm{G}\right)=6.67\times 10-11\mathrm{m}3\mathrm{kg}-1\mathrm{s}-2$
Mass of the Earth $\left(\mathrm{M}\right)=5.97\times 1024\mathrm{kg}$

Step 2: Formula used

${\mathrm{m\omega }}^2(\mathrm{R}+\mathrm{h})=\mathrm{GMem}/(\mathrm{R}+\mathrm{h}{)}^2$

$\mathrm{V}=\mathrm{\omega }\times (\mathrm{R}+\mathrm{h})$

$\mathrm{T}=24\mathrm{hrs}=24\times 60\times 60\mathrm{s}=86400\mathrm{s}$

where,

$\mathrm{V}$ is the velocity of the satellite.

$\mathrm{T}$ is the time taken.

Step 3: Calculation of height from the earth

Let the length of such orbit out from Earth's surface be $\text{'}\mathrm{h}\text{'}\mathrm{m}$, and a satellite of mass $\text{'}\mathrm{m}\text{'}\mathrm{kg}$ swings inside this earth orbit with a homogenous rotational velocity of $\mathrm{rad}/\mathrm{s}.$

$(2\mathrm{\pi }/\mathrm{T}{)}^2(\mathrm{R}+\mathrm{h})=\mathrm{GMe}(\mathrm{R}+\mathrm{h})2$

$(\mathrm{R}+\mathrm{h}{)}^3={\mathrm{GMeT}}^2/4{\mathrm{\pi }}^2$

$\mathrm{h}=42241-\mathrm{R}$

$=42241-6380$

$=35861\mathrm{km}$

Step 4: Find the satellite velocity

$\mathrm{v}=\mathrm{\omega }\times (\mathrm{R}+\mathrm{h})$

$=2\mathrm{\pi }/\mathrm{T}\times 4224.1\times 104$

$\approx 3070\mathrm{m}/\mathrm{s}$

Therefore, the height from the Earth's surface is $3070\mathrm{m}/\mathrm{s}$.