The earth is assumed to be a sphere of radius $R$. A platform is arranged at a height $R$ from the surface of

Question

The earth is assumed to be a sphere of radius $R$. A platform is arranged at a height $R$ from the surface of the earth. The escape velocity of a body from this platform is $fv_e$, where $v_e$ is its escape velocity from the surface of the earth. The value of $f$ is:

Accepted Answer

1. **Escape Velocity Formula:** The escape velocity $v_e$ from the surface of the Earth (radius $R$) is given by $v_e = \sqrt{\frac{2GM}{R}}$.
2. **Escape Velocity at Altitude:** The escape velocity from a point at a distance $r$ from the center of the Earth is $v_{esc} = \sqrt{\frac{2GM}{r}}$.
3. **Given Condition:** The platform is at a height $h = R$ from the surface. Therefore, the distance from the center is $r = R + h = R + R = 2R$.
4. **Calculation:** Substitute $r = 2R$ into the escape velocity formula:
   $$ v_{platform} = \sqrt{\frac{2GM}{2R}} = \frac{1}{\sqrt{2}} \sqrt{\frac{2GM}{R}} $$
   Since $v_e = \sqrt{\frac{2GM}{R}}$, we have:
   $$ v_{platform} = \frac{1}{\sqrt{2}} v_e $$
5. **Finding f:** Comparing this with the given expression $v_{platform} = f v_e$, we find:
   $$ f = \frac{1}{\sqrt{2}} $$

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