The energy required to take a satellite to a height ' $h$ ' above Earth surface (radius of Earth $=6.4 \times 10^{3} \mathrm{~km}$) is $\mathrm{E}_{1}$ and kinetic energy required for the satellite to be in a circular orbit at this height is $\mathrm{E}_{2}$. The value of h for which $\mathrm{E}_{1}$ and $\mathrm{E}_{2}$ are equal is

(B) $3.2 \times 10^{3} \mathrm{~km}$

(A) $1.6 \times 10^{3} \mathrm{~km}$

(C) $6.4 \times 10^{3} \mathrm{~km}$

(D) $1.28 \times 10^{4} \mathrm{~km}$

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Gravitation

The energy required to take a satellite to a height '

h

' above Earth surface (radius of Earth

= 6.4 \times 1 0^{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

Answer: (B)

Lifting the satellite from the surface to height

h

changes only its potential energy, so

E_{1} = (- \frac{GM m}{R + h}) - (- \frac{GM m}{R}) = \frac{GM m h}{R ( R + h )}

. Once there, a circular orbit needs orbital speed

v

with

\frac{m v ^{2}}{R + h} = \frac{GM m}{( R + h ) ^{2}}

, so the orbital kinetic energy is

E_{2} = \frac{1}{2} m v^{2} = \frac{GM m}{2 ( R + h )}

. Setting

E_{1} = E_{2}

, the common factor

\frac{GM m}{R + h}

cancels and

\frac{h}{R} = \frac{1}{2}

, i.e.

h = \frac{R}{2} = \frac{6.4 \times 1 0 ^{3}}{2} = 3.2 \times 1 0^{3} km

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option