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The energy required to take a satellite to a height 'h' above Earth surface (radius of the Earth = 6.4 X 10³km) is E₁ and kinetic energy required for satellite to be in a circular orbit at this height is E₂. The value of h for which E₁ and E₂ are equal, is:

Option 1)
1.6 X 10³km
Option 2)
6.4 X 10³km
Option 3)
1.28 X 10⁴km
Option 4)

3.2 X 10³km

Answers (1)

best_answer

Energy of satellite -

$K=\frac{1}{2}mV^{2}=\frac{GMm}{2r}$

$U=-\frac{GMm}{r}$

$K\rightarrow Kinetic\: energy$

$U\rightarrow Potential\: energy$

$M\rightarrow mass\: of\: planet$

$m\rightarrow mass\: of\: satellite$

$E=K+U$

$E=-\frac{GMm}{2r}$

$E=Total\: energy$

- wherein

$K=-E$

$U=2E$

$U=-2K$

$V_{Surfuse}+E_{1}=U_{h}$

$\frac{-GM_{e}m}{R_{e}}+E_{1}=\frac{-GM_{e}m}{R_{e}+h}$

$E_{1}=GM_{e}m (\frac{1}{R_{e}}-\frac{1}{R_{e}+h})$

$E_{1}=\frac{GM_{e}m}{R_{e}+h}.\frac{h}{R_{e}}$

Gravitational force

$F=ma_{c}=\frac{mv^{2}}{R_{e}+h}$

$E_{2}\Rightarrow \frac{mv^{2}}{R_{e}+h}=\frac{GM_{e}m}{(R_{e}+h)^{2}}$

$mv^{2}=\frac{GM_{e}m}{R_{e}+h}$

$E_{2}=\frac{mv^{2}}{2}=\frac{GM_{e}m}{2(R_{e}+h)}$

$E_{1}=E_{2}$

$\therefore \frac{h}{R_{e}}=\frac{1}{2}\Rightarrow h=\frac{R_{e}}{2}$

$=3200\: km$

Option 1)

1.6 X 10³km

Option 2)

6.4 X 10³km

Option 3)

1.28 X 10⁴km

Option 4)

3.2 X 10³km

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