Question

The minimum energy required to launch a m kg satellite from earth surface in a circular orbit at a height 2 r will be

Answer 1

Answer on Question #41178 - Physics - Other

Question.

The minimum energy required to launch a m kg satellite from earth surface in a circular orbit at a height 2r will be

Solution.

We proceed in non-inertial reference frame with respect to the Earth. In this case the object in orbit will be at rest, as two forces act on him: the centrifugal force and the gravitational force.

Gravitational force:

$$F = G \frac {M m}{R ^ {2}}$$

$G$ is a gravitational constant; $G = 6.67 \cdot 10^{-11} \frac{N \cdot m^2}{kg^2}$

$M$ is the mass of Earth; $M = 6 \cdot 10^{24} kg$

$m$ is the mass of satellite;

$R$ is a distance between the center of the Earth and the satellite in orbit;

Centrifugal force:

$$F = \frac {m v ^ {2}}{R}$$

Equate these formulas and calculate the square of rate $v^2$:

$$G \frac {M m}{R ^ {2}} = \frac {m v ^ {2}}{R}$$ $$v ^ {2} = \frac {G M}{R}$$

Here, $R = R_{E} + 2r$, where

$R_{E}$ is a radius of Earth; $R_{E} = 6400 \, \text{km}$

$2r$ is a height of orbit

The minimum energy required to launch a satellite is

$$E = \frac {m v ^ {2}}{2}$$

So,

$$E = \frac {G M m}{2 R} = \frac {G M m}{2 (R _ {E} + 2 r)}$$

Answer.

$$E = \frac {G M m}{2 (R _ {E} + 2 r)}$$

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