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Step-by-Step Solution
Step 1: Understand the Problem
We have the Earth modeled as a sphere of radius $R$. A platform is placed at a height $R$ above the Earth’s surface, so the distance from the center of the Earth to the platform is $2R$. We want to find the escape velocity from this platform in terms of the known escape velocity $v$ from the Earth’s surface.
Step 2: Recall the Escape Velocity from the Earth's Surface
The escape velocity $v$ from the surface of the Earth (radius $R$) is given by:
$ v = \sqrt{2gR} $,
where $g$ is the acceleration due to gravity at the Earth’s surface.
Step 3: Express the Escape Velocity at a Height $R$ above the Surface
If the height above the Earth’s surface is $R$, the radial distance from the center is $2R$. The escape velocity $v_{\text{platform}}$ from a distance $2R$ (assuming gravitational potential energy is zero at infinity) satisfies:
$ \tfrac{1}{2} m\,v_{\text{platform}}^{2} - \tfrac{GMm}{2R} = 0, $
since the potential energy at $2R$ is $- \tfrac{GMm}{2R}$ and we want the total energy (kinetic + potential) to equal 0 for just escape. Here, $G$ is the universal gravitational constant and $M$ is the mass of the Earth. Rearranging,
$ \tfrac{1}{2} m\,v_{\text{platform}}^{2} = \tfrac{GMm}{2R}. $
Dividing both sides by $m/2$ gives:
$ v_{\text{platform}}^{2} = \tfrac{GM}{R}. $
Thus,
$ v_{\text{platform}} = \sqrt{\tfrac{GM}{R}}. $
Step 4: Compare with Escape Velocity from the Surface
From the surface, we know $ v = \sqrt{2gR} $. Also, near the Earth’s surface $ g = \tfrac{GM}{R^2} $. Hence,
$ \sqrt{\tfrac{GM}{R}} = \sqrt{\tfrac{GM}{R^2} \cdot R} = \sqrt{g R}.
$
So,
$ v_{\text{platform}} = \sqrt{gR}.
$
Step 5: Determine the Factor $f$
We want to express $ v_{\text{platform}} $ in terms of $ v $. Since $ v = \sqrt{2gR} $, we see that:
$ \sqrt{gR} = \tfrac{1}{\sqrt{2}} \sqrt{2gR} = \tfrac{v}{\sqrt{2}}.
$
Therefore,
$ v_{\text{platform}} = \tfrac{v}{\sqrt{2}}.
$
Hence, if $ v_{\text{platform}} = f \, v $, then
$ f = \tfrac{1}{\sqrt{2}}.
$
Final Answer
The value of $ f $ is $ \tfrac{1}{\sqrt{2}} $.
