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While establishing relation of escape velocity and radius , I confronted a problem .

(i) $$v_e=\sqrt{\frac{2GM}{R}}$$

This states that $v_e$ is inversely proportional to the square root of radius .

(i) $$v_e=\sqrt{2gR}$$

This states that $v_e$ is directly proportional to the square root of radius .

Whoch one’s correct?

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  • $\begingroup$ The two formulas do not contradict, because they do not use the same variables. $\endgroup$ Mar 10, 2018 at 22:14

4 Answers 4

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The problem is solved when you realize that $g$ in your second formula is a function of $R$: if

$$F = \frac{GMm}{r^2}$$

is rewritten as

$$F = mg$$

Then it follows that

$$g = \frac{GM}{r^2}$$

When you substitute that in your second equation, you get the first...

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  • $\begingroup$ That means the correct relation is obtained from the first one .. $\endgroup$
    – user179960
    Mar 8, 2018 at 13:13
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    $\begingroup$ They are both correct if your second one has the square root for the whole thing. $\endgroup$ Mar 8, 2018 at 13:15
  • $\begingroup$ How come ...? Isnt that condradictory? $\endgroup$
    – user179960
    Mar 8, 2018 at 13:18
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    $\begingroup$ It is not contradictory. If you found yourself in a very high orbit (say out near the moon), you could escape the earth's gravity from there with a smaller velocity than you would need close to earth. The second equation you have expresses things relative to the local gravity at the height where you are (which itself drops off with inverse square of distance to center of earth). So rather than considering $g$ constant and getting confused, you should recognize it's really $g(r)$, expand the expression, and realize there is no contradiction. $\endgroup$
    – Floris
    Mar 8, 2018 at 13:27
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    $\begingroup$ Thank you ... that is , the g in the second one is a function of R ... there are two variables ... $\endgroup$
    – user179960
    Mar 8, 2018 at 13:29
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When you write a relationship between two variables $x$ and $y$ as $y \propto x$ you can also write this as $y=kx$ where $k$ is a constant independent of $x$ and $y$.

Assume that $R$ is the radius of the planet, $M$.

Using your first equation you stated that the escape velocity $v_{\rm c}$ is inversely proportional to the square root of the radius, $\sqrt R$, provided that the mass, $M$, is constant.
But you have a problem because assuming that it is a homogeneous spherical planet of density $\rho$ the mass $M = \frac 43 \pi R^3 \rho$ and the mass depends on the radius so what you assumed to be a constant, $2GM$, is not independent of the radius.

For the second equation you are saying that the escape velocity $v_{\rm c}$ is proportional to the square root of the radius, $\sqrt R$, provided $g$ is constant.
However $g = \frac{GM}{R^2}$ so it is in fact also dependent on $R$ thus invalidating your second proportionality.

You can however show that if the density is constant then the escape velocity from a homogeneous spherical planet is proportional to the radius of the planet.

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You can derive the scape velocity by calculating the energy at the surface and then at infinity

$$ E_{\rm surf} = \frac{1}{2}mv^2 -G\frac{M m}{R} \tag{1} $$

You want to find $v$ such that the mass $m$ reaches infinity $r\to\infty$ with zero velocity, that is

$$ E_{\inf} = 0 \tag{2} $$

If you put these two equations together

$$ \frac{1}{2}mv_e^2 - G\frac{Mm}{R} = 0 ~~~\Rightarrow~~~ v_e = \left(\frac{2GM}{R}\right)^{1/2} \tag{3} $$

But now, the force that $m$ feels on the surface is

$$ F = -G\frac{Mm}{R^2} = -\left(\frac{G M}{R^2}\right)m = -g m ~~~\mbox{with}~~~ g = \frac{GM}{R^2}\tag{4} $$

Replacing Eq. (4) in Eq. (3) you get

$$ v_e = \left(2 g R\right)^{1/2} \tag{5} $$

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  • $\begingroup$ Would you please kindly state the correct relation between g and R ? $\endgroup$
    – user179960
    Mar 8, 2018 at 13:15
  • $\begingroup$ @A.H.M. Please look at the last part of Eq. (4) $\endgroup$
    – caverac
    Mar 8, 2018 at 13:29
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$g$ is defined as: $$g=\frac{GM}{R^2}$$

The escape velocity is:

$$v_{e}=\left(\frac{2GM}{R}\right)^{0.5}$$

You can write this equation in terms of $g$ doing this trick:

$$v_{e}=\left(\frac{2GM}{R}\frac{R}{R}\right)^{0.5}=\left(2\frac{GM}{R^2}R\right)^{0.5}=\left(2gR\right)^{0.5}$$

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