Question

A remote sensing satellite is moving around the earth in an orbit whose altitude from the surface of earth is 800km calculate orbital velocity and orbital period of the satellite​

Answer 1

Answer:

Orbital velocity = 7.455 × 10^6 m/s

Time Period T = 1.69 hour

Explanation:

Taking the following data

Radius of earth Re = 6400 km

Mass of the earth Me = 6 × 10^24 kg

G = 6.67 × 10^-11 N-m²/kg²

Given

Height of the satellite h = 800 km

R = Re + h = 6400 + 800 = 7200 km = 7.2 × 10^6 m

Orbital velocity of the satellite is given by

$v=\sqrt{\frac{GMe}{R}}$

$\implies v=\sqrt{\frac{6.67\times 10^{-11\times}6\times10^{24}}{7.2\times 10^6}}$

$\implies v=7.455\times 10^3 \text{ m/s}$

Calculation of Time period

using

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

we get

$T^2=\frac{4\pi^2(7.2\times10^6)^3}{6.67\times 10^{-11}\times 6\times 10^{24}}$

$\implies T=2\pi\times\sqrt{9.3265\times10^5}$

$\implies T=2\pi\times 9.657\times10^2$

$\implies T=6067.67 \text{ sec}$

or, T = 1.69 hour