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

Question

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

Infinity Learn · Accepted Answer

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

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 $g = 9.8 {ms}^{- 2}$ , then the orbital speed of the satellite is

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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

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

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 $g = 9.8 {ms}^{- 2}$ , then the orbital speed of the satellite is