The energy to take a satellite to height h above Earth (radius R_E = 6.4 times 10^3 text km ) is E_1, and the kinetic energy to maintain circular orbit at that height is E_2. For what value of h is E_1 = E_2?

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EEJ MAIN Physics QUESTION #7002

Question 1

The energy to take a satellite to height $h$ above Earth (radius $R_E = 6.4 \times 10^3\ \text{km}$) is $E_1$, and the kinetic energy to maintain circular orbit at that height is $E_2$. For what value of $h$ is $E_1 = E_2$?

$1.6 \times 10^3\ \text{km}$
$3.2 \times 10^3\ \text{km}$✔️
$6.4 \times 10^3\ \text{km}$
$1.28 \times 10^4\ \text{km}$

Correct Answer Explanation

$E_1 = \frac{GMm}{R_E} - \frac{GMm}{R_E+h} = GMm\frac{h}{R_E(R_E+h)}$

$E_2 = \frac{GMm}{2(R_E+h)}$ (orbital KE)

Setting $E_1 = E_2$:

$\frac{h}{R_E(R_E+h)} = \frac{1}{2(R_E+h)}$

$\frac{h}{R_E} = \frac{1}{2}$

$h = \frac{R_E}{2} = \frac{6.4 \times 10^3}{2} = 3.2 \times 10^3\ \text{km}$

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