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Byju's Answer

Standard XII

Physics

Kepler's Law

A 400 kg sate...

Question

A $400 k g$ satellite is placed in earth's orbit at $3600 k m$ above the surface. What is the orbital speed of the satellite?
$(R = 6400 k m) (g_{o} = 10 m / s^{2})$

6.4 k m / s

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4.8 k m / s e c

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7.6 k m / s

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8.4 k m / s

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Solution

The correct option is A $6.4 k m / s$

Orbital speed of a satellite is given as:
$V_{0} = \sqrt{\frac{G M}{r}} = \sqrt{\frac{G M}{R^{2}} . \frac{R^{2}}{r}} = \sqrt{\frac{g_{o} R^{2}}{r}} = R \sqrt{\frac{g_{o}}{r}} = 6400 \sqrt{\frac{10 \times 10^{- 3}}{10000}} = 6400 / 1000 = 6.4 k m / s$

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A 400 kg satellite is placed in earth's orbit at 3600 km above the surface. What is the orbital speed of the satellite? (R=6400 km) (go=10 m/s2)

The correct option is A 6.4 km/s Orbital speed of a satellite is given as: V0=√GMr=√GMR2.R2r=√goR2r=R√gor=6400√10×10−310000=6400/1000=6.4 km/s

A $400 k g$ satellite is placed in earth's orbit at $3600 k m$ above the surface. What is the orbital speed of the satellite?
$(R = 6400 k m) (g_{o} = 10 m / s^{2})$

The correct option is A $6.4 k m / s$

Orbital speed of a satellite is given as:
$V_{0} = \sqrt{\frac{G M}{r}} = \sqrt{\frac{G M}{R^{2}} . \frac{R^{2}}{r}} = \sqrt{\frac{g_{o} R^{2}}{r}} = R \sqrt{\frac{g_{o}}{r}} = 6400 \sqrt{\frac{10 \times 10^{- 3}}{10000}} = 6400 / 1000 = 6.4 k m / s$