JEE PYQ: Gravitation Question 42
Question 42 - 2019 (09 Jan Shift 2)
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) $1.6 \times 10^3$ km
(2) $3.2 \times 10^3$ km
(3) $6.4 \times 10^3$ km
(4) $28 \times 10^4$ km
Show Answer
Answer: (2)
Solution
$E_1 = \frac{GM_e m}{R_e} \cdot \frac{h}{R_e + h}$ and $E_2 = \frac{GM_e m}{2(R_e+h)}$. Setting $E_1 = E_2$: $\frac{h}{R_e} = \frac{1}{2}$, so $h = R_e/2 = 3200$ km.
BETA
