Chapter 13: Q. 30 (page 354)

An earth satellite moves in a circular orbit at a speed of $5500 m / s$ . What is its orbital period?

Short Answer

Expert verified

The orbital period is $4.18 h o u r s .$

Step by step solution

Given information

Speed of orbiting = $4.18 h o u r s .$

Calculation

The centripetal acceleration required for circular motion of satellite is provided by the gravitational force between the satellite and the earth, therefore : $G M_{e} m_{s} / r_{s}^{2} = m_{s} v_{s}^{2} / r_{s}$

Here, $M_{e}$ is the mass of earth, $m_{s}$ is the mass of satellite, G is the gravitational constant, r is the radius of the orbit.

Now solving we get,

$G M_{e} m_{s} / r_{s}^{2} = m_{s} v_{s}^{2} / r_{s} r_{s} = G M_{e} / v_{s}^{2} r_{s} = \frac{(6.67 \times 10^{- 11} N m^{2} / k g^{2}) (5.98 \times 10^{24} k g)}{{(5500 m / s)}^{2}} r = 13.18 \times 10^{6} m$

Orbital period T is given by according to Kepler's Law :

$T = 2 π \sqrt{\frac{r_{s}^{3}}{G M_{e}}} T = 2 π \sqrt{\frac{{(13.18 \times 10^{6} m)}^{3}}{(6.67 \times 10^{- 11} N m^{2} / k g^{2}) (5.98 \times 10^{24} k g)}} T = 2 π \sqrt{5740096.7 s^{2}} T = 2 π \times 2395.84 s T = 15053 \sec T = 4.18 hours$

Therefore, time period = $4.18 h o u r s .$

Final answer

The orbital period is $4.18 h o u r s .$

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An earth satellite moves in a circular orbit at a speed of $5500 m / s$ . What is its orbital period?

Short Answer

Step by step solution

Given information

Calculation

Final answer

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An earth satellite moves in a circular orbit at a speed of 5500m/s. What is its orbital period?

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Step by step solution

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Calculation

Final answer

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An earth satellite moves in a circular orbit at a speed of $5500 m / s$ . What is its orbital period?