Question

A satellite orbits around the Earth at a height equal to R of the Earth. Find its period.
(R = 6.4 x 10⁶ m, g = 9.8 m/s?)
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Answer 1

$\huge{\underline{\underline{\bf{ SoLuTiOn\: - }}}}♡$

Mass of the Earth , M = 6.0 × 1024 kg

m = 200 kg

 Re = 6.4 × 106 m

G = 6.67 × 10 − 11 Nm 2kg − 2

Height of the satellite , h = 400 km = 4 × 105 m

Total energy of the satellite at height h=(1/2)mv2+(−G(Re+h)Mem)

Orbital velocity of the satellite, v=Re+hGMe

Total energy at height h =21Re+hGMem−Re+hGMem

Total Energy=−21Re+hGMem

The negative sign indicates that the satellite is bound to the Earth. 

Energy required to send the satellite out of its orbit = – (Bound energy)

=2(Re+h)GMem

=2(6.4×106+4×105)6.67×10−11×6×1024×200

=5.9×109 J

If the satellite just escapes from the gravitational field, then total energy of the satellite is zero. Therefore, we have to supply 5.9×109J of energy to just escape it.

Answer 2

Given :A satellite orbits around the Earth at a height equal to R of the Earth. Find its period.

(R = 6.4 x 10⁶ m, g = 9.8 m/s?)

R = 6.4 * 10 ⁶ m

g = 9.8 m/ s

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Formula used

T =

$\frac{2\pi( R+ h)}{v0}$

v0 = √ Gm / R + h

√6.67 * 10 -¹¹ * 6 * 10 ²⁴ / 2 * 6.4 * 10 ⁶

T =

$\frac{2 \times 3.14 \times 2 \times 6.4 \times 10}{5.592}$

4 hrs

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So the answer is 4 hrs .

Formula used :-

1. T = $\frac{2\pi( R+ h)}{v0}$

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We count the height of Earth with the help of h then put the formula .