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NEET PHYSICSMedium

The earth is assumed to be a sphere of radius RR. A platform is arranged at a height RR from the surface of the earth. The escape velocity of a body from this platform is fvefv_e, where vev_e is its escape velocity from the surface of the earth. The value of ff is:

A

2\sqrt{2}

B

12\frac{1}{\sqrt{2}}

C

13\frac{1}{3}

D

12\frac{1}{2}

Step-by-Step Solution

  1. Escape Velocity Formula: The escape velocity vev_e from the surface of the Earth (radius RR) is given by ve=2GMRv_e = \sqrt{\frac{2GM}{R}}.
  2. Escape Velocity at Altitude: The escape velocity from a point at a distance rr from the center of the Earth is vesc=2GMrv_{esc} = \sqrt{\frac{2GM}{r}}.
  3. Given Condition: The platform is at a height h=Rh = R from the surface. Therefore, the distance from the center is r=R+h=R+R=2Rr = R + h = R + R = 2R.
  4. Calculation: Substitute r=2Rr = 2R into the escape velocity formula: vplatform=2GM2R=122GMRv_{platform} = \sqrt{\frac{2GM}{2R}} = \frac{1}{\sqrt{2}} \sqrt{\frac{2GM}{R}} Since ve=2GMRv_e = \sqrt{\frac{2GM}{R}}, we have: vplatform=12vev_{platform} = \frac{1}{\sqrt{2}} v_e
  5. Finding f: Comparing this with the given expression vplatform=fvev_{platform} = f v_e, we find: f=12f = \frac{1}{\sqrt{2}}
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