ExamBro
ExamBro
JEE Mains · Physics · STD 11 - 7. gravitation

The energy required to take a satellite to a height \(‘h’\) above Earth surface (radius of Earth \(= 6.4 \times 10^3\,km\) ) is \(E_1\) and kinetic energy required for the satellite to be in a circular orbit at this height is \(E_2.\) The value of \(h\) for which \(E_1\) and \(E_2\) are equal is

  1. A \(1.6\times 10^3\,km\)
  2. B \(3.2\times 10^3\,km\)
  3. C \(6.4\times 10^3\,km\)
  4. D \(1.28\times 10^4\,km\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3.2\times 10^3\,km\)

Step-by-step Solution

Detailed explanation

\({E_1} =-\frac{{GMm}}{{R + h}} - \left( { - \frac{{GMm}}{R}} \right)\) \({E_2} = \frac{1}{2}m{\left( {\sqrt {\frac{{GM}}{{R + h}}} } \right)^2} = \frac{{GMm}}{{2\left( {R + h} \right)}}\) \({E_1} = {E_2}\,\,;\,\,h = \frac{R}{2}\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app