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A satellite revolving around the Earth in a circular orbit of radius r, is brought to a smaller circular orbit of radius $(r - ∆ r)$ . If $T_{0}, v_{0}$ and E0 are in the original period, velocity and modulus value of total energy, then after the change in radius : $(∆ r < < r)$

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period decreases by $ΔT = T_{0} \frac{3}{2} \frac{Δr}{r}$

period increases by $ΔT = T_{0} \frac{3}{2} \frac{Δr}{r}$

velocity increases by $Δv = v_{0} \frac{1}{2} \frac{Δr}{r}$

total energy decreases by $ΔE = E_{0} \frac{1}{2} \frac{Δr}{r}$

answer is A, C, D.

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Detailed Solution

$T = \frac{2 π}{\sqrt{GM}} r^{3 / 2}; v = \sqrt{\frac{GM}{r}}; E = - \frac{GMm}{r}$

$\Rightarrow T \propto r^{3 / 2}; v \propto r^{- 1 / 2}; E \propto r^{- 1}$

[use the result $A \propto x^{n} \Rightarrow \frac{δ A}{A} = n \frac{δ x}{x}]$

$\frac{δ T}{T} = (\frac{3}{2}) \frac{δ r}{r}; \frac{δ v}{v} = (- \frac{1}{2}) \frac{δ r}{r}; \frac{δ E}{E} = (- 1) \frac{δ r}{r}$

Put $T = T_{0}; v = v_{0}; E = - E_{0}$ (since E is negative and $E_{0}$ stands for its modulus).

Also put $δ r = - Δr$ we get A,C and correct.

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