Physics, asked by ashwaniverma1165, 10 months ago

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

Answers

Answered by rajib30
0

Answer:

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Answered by CarliReifsteck
0

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}}

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