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A spacecraft is at 4252km above the earth's surface.if it descends at the rate of 5km per minute.what will be its position after 9 hours​

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Position of the Spacecraft = 4252 km above the earh surface
In 9 hrs it will descends = 9 x 60 x 5 = 2700 km
Position of the spacecraft after 9 hrs = 4252 - 2700=  1552 km above the earth

Posted by

Ravindra Pindel

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