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Question

# An artificial satellite is moving in a circular orbit around the earth with a speed equal to 38 times the magnitude of the escape velocity from the earth. If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth, then the speed with which it hits the surface of the earth is √23n km/s. Find n. (take g = 10 m/s and Re = 6400 Km)

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Solution

## Escape velocity ve=√2gRe Given, speed of the satelliitev=38√2gRe=√9gRe32 --- (i) Orbital velocity at height ′h′=√GMRe+h =√GM.R2eR2e(Re+h)=√R2egRe+h ---(ii) Equating (i) and (ii), √9gRe32=√R2egRe+h ⇒9Re+9h=32Re ⇒h=239Re Now, total energy at height ′h′ = total energy at the surface of the earth. 0−GMemRe=12mv2−GMemRe+h 12mv2=GMemRe−GMemRe+239Re 12mv2=2332GMemRe ⇒v=√2316GMeRe ⇒v=√2316gRe=√232 km/s (∵ √gRe=8 km/sec) ∴n=2

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