Question

A remote sensing satellite is movig around the earth in an orbit whose altitude from the surface of earth is 800km . Calculate orbital velocity and oebital period of the satellite

Answer 1

this orbiting satellite is given by the relationship

Orbital Period Equation

The final equation that is useful in describing the motion of satellites is Newton's form of Kepler's third law. Since the logic behind the development of the equation has been presented elsewhere, only the equation will be presented here.

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Answer 2

Answer:

Orbital velocity = 7.455 × 10^6 m/s

Time Period T = 1.69 hour

Explanation:

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

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

V= √ GMe

          R

= V= √ 6.67*10−¹¹ * 6* 10−²⁴

               7.2*10⁶

= V = 7.455*10³ m/s

Calculation of time period using

T² = 4π² (7.2*10⁶)³

          6.67*10−¹¹ *6*10−²⁴

T = 2π * √9.3265*10⁵

T = 2π * 9.657*10²

T = 6067.67sec OR T = 1.69hour