A remote - sensing satellite of earth revolves in a circular orbit at a height of 0-25 - 106 m above the surface of earth- If earth-s radius is 6-38- 106m and g - 9-8 ms-2- then the orbital speed of the satellite is-

The correct option is D 7.76 km s^ 1 Orbital speed: V = √(GM/r)= √(GM/R^2R^2/r)=√(gR^2/r)= √(9.8 × ( 6.38 × 10^6)^2/ 0.25 × 10^6)=7.76 km/s ...

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Standard XII

Physics

Centripetal Force

A remote - se...

Question

A remote - sensing satellite of earth revolves in a circular orbit at a height of $0.25 \times 10^{6}$ m above the surface of earth. If earth's radius is $6.38 \times 10^{6} m$ and $g = 9.8 m s^{- 2}$ , then the orbital speed of the satellite is:

6.67 k m s^{- 1}

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7.76 k m s^{- 1}

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8.56 k m s^{- 1}

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9.13 k m s^{- 1}

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Solution

The correct option is D $7.76 k m s^{- 1}$
Orbital speed: $V = \sqrt{\frac{G M}{r}} = \sqrt{\frac{G M}{R^{2}} \frac{R^{2}}{r}} = \sqrt{\frac{g R^{2}}{r}} = \sqrt{\frac{9.8 \times (6.38 \times 10^{6})^{2}}{0.25 \times 10^{6}}} = 7.76 k m / s$

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A remote - sensing satellite of earth revolves in a circular orbit at a height of 0.25×106 m above the surface of earth. If earth's radius is 6.38×106m and g=9.8ms−2, then the orbital speed of the satellite is:

The correct option is D 7.76kms−1Orbital speed: V=√GMr=√GMR2R2r=√gR2r=√9.8×(6.38×106)20.25×106=7.76km/s

A remote - sensing satellite of earth revolves in a circular orbit at a height of $0.25 \times 10^{6}$ m above the surface of earth. If earth's radius is $6.38 \times 10^{6} m$ and $g = 9.8 m s^{- 2}$ , then the orbital speed of the satellite is:

The correct option is D $7.76 k m s^{- 1}$
Orbital speed: $V = \sqrt{\frac{G M}{r}} = \sqrt{\frac{G M}{R^{2}} \frac{R^{2}}{r}} = \sqrt{\frac{g R^{2}}{r}} = \sqrt{\frac{9.8 \times (6.38 \times 10^{6})^{2}}{0.25 \times 10^{6}}} = 7.76 k m / s$