Q.

A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25×106m 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

Moderate

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9.13 km s-1

6.67 km s-1

7.76 km s-1

8.56 km s-1

answer is C.

(Detailed Solution Below)

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Detailed Solution

The orbital speed of the satellite is vo=Rg(R+h)where R is the earth's radius, g is the acceleration due to gravity on earth's surface and h is the height above the surface of earth. Here, R=6.38×106 m,g=9.8 m s-2and h=0.25×106 m∴vo=6.38×106 m9.8 m s-26.38×106 m+0.25×106 m=7.76×103 m s-1=7.76 km s-1∵1 km=103 m

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