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MHT CET · Physics · Gravitation

The gravitational potential energy required to raise a satellite of mass ' \(m\) ' to height ' \(h\) ' above the earth's surface is ' \(E_1\) '. Let the energy required to put this satellite into the orbit at the same height be ' \(E_2\) '. If \(M\) and \(R\) are the mass and radius of the earth respectively then \(E_1: E_2\) is

  1. A \(\mathrm{h}: \mathrm{R}\)
  2. B \(h: 2 R\)
  3. C \(\mathrm{R}: \mathrm{h}\)
  4. D \(2 \mathrm{~h}: \mathrm{R}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(2 \mathrm{~h}: \mathrm{R}\)

Step-by-step Solution

Detailed explanation

Energy required to raise a satellite upto a height \(h\),
\(\mathrm{E}_1=\Delta \mathrm{U}=\frac{\mathrm{mgh}}{1+\frac{\mathrm{h}}{\mathrm{R}}}\)...(i)
Energy required to put satellite into orbit, \(\mathrm{E}_2=\frac{1}{2} \mathrm{mv}_0^2=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{GM}}{\mathrm{r}}\right) \quad\) as \(\mathrm{v}_0\) is orbital speed
\(=\frac{1}{2} m\left(\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}\right)\)
Dividing numerator and denominator by \(\mathrm{R}^2\),
\(\begin{aligned}
E_2 & =\frac{1}{2} m\left(\frac{\frac{G M}{R^2}}{\frac{R+h}{R} \times \frac{1}{R}}\right) \\
& =\frac{1}{2} m\left(\frac{g}{1+\frac{h}{R}}\right) R \\
E_2 & =\frac{m g R}{2\left(1+\frac{h}{R}\right)} \quad \ldots\left(\because g=\frac{G M}{R^2}\right) \\
\frac{E_1}{E_2} & =\frac{2 h}{R}
\end{aligned}\)
\(\ldots[\) From(i) and (ii) \(]\)