Question

A satellite is rotating very close to Earth's surface has approximately linear speed equal to

Answer 1

Answer:

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Answer 2

Given that,

A satellite is rotating very close to Earth's surface has approximately

Let the satellite of mass m rotating around a earth of mass M in circular orbit of radius r.

We need to calculate the linear speed of earth's surface

centripetal force is provided by gravitational attractive force

$\dfrac{mv^2}{r}=\dfrac{GmM}{r^2}$

Where, r = radius

G = gravitational constant

M = mass of earth

m = mass of satellite

Put the value into the formula

$v^2=\dfrac{GM}{r}$

$v=\sqrt{\dfrac{GM}{r}}$

Hence, The linear speed of earth's surface is equal to $\sqrt{\dfrac{GM}{r}}$