Question

A satellite is revolving around the earth in an orbit of radius r with time period T . If the satellite is revolving around the earth in an orbit of radius r + ∆r (∆r<<r) with time period T + ∆T(∆T<<T), then​

Answer 1

Answer:

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Explanation:

By the way answer is ∆T/T=3/2∆r/r (1st Option)

Answer 2

Initially, the satellite was revolving in an orbit of radius R in time interval T. Now, the same satellite is revolving around in an orbit of radius R + ∆R in a time interval T + ∆T.

From Kepler's Third Law,

$\sf (Time \: lnterval) {}^{2} \propto {Radius \: of \: orbit} {}^{3} \\ \\ \longrightarrow \sf T {}^{2} = k R {}^{3}$

Differentiating the above expression,

$\longrightarrow \sf \: 2 \dfrac{\Delta{T} }{T} =3\dfrac{\Delta{R}}{R} \\ \\ \longrightarrow \boxed{ \boxed{\sf \:\dfrac{\Delta{T} }{T} =\dfrac{3}{2} \dfrac{\Delta{R}}{R}}}</p><p></p><p>$

Option (A) is correct.