Question

An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth. If the satellite is stopped in its orbit and allowed to fall freely onto the earth, the speed with which it hits the surface ______ km/s.

[g = 9.8 ms^-2 and R_e = 6400 km]

विकल्प

• 7.919

• 8.200

• 6.123

• 5.478

Answer 1

An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth. If the satellite is stopped in its orbit and allowed to fall freely onto the earth, the speed with which it hits the surface is 7.919 km/s.

Explanation:

Let the mass of earth = M

Radius = R

Then escape velocity from the earth's surface

`V_e = sqrt(2gR)`

Velocity of satellite `V_s = V_e/2 = sqrt(2gR)/2` ..........(i)

`V_s = sqrt((GM)/r) = sqrt((R^2g)/(R + h))`

`V_s^2 = sqrt((R^2g)/(R + h)` .............(ii)

From equations (i) and (ii),

We get h = R = 6400 km

Now total energy at height h = total energy at the earth's surface

∴ `0 - (GMm)/(R + h) = 1/2mV^2 - (GMm)/R`

Solving we get V = `sqrt(R/g)`

V = `sqrt(64000 xx 10^3 xx 9.8)`

V = 7.92 km/sec