A man can jump vertically to a height of $1.5 m$ on the earth. Calculate the radius of a planet of the same mean density as that of the earth from whose gravitational field he could escape by jumping. Radius of earth is $6.41 \times 1 0^{6} m$ .

Question

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Answer 1

$h = \frac{u ^{2}}{2 g _{e}}$
$∴$    $u = 2 g_{e} h$           (i)
For the asked planet this u sbould be equal to the escape velocity from its surface.
$2 g_{e} h = 2 g_{p} R_{p}$
or $g_{e} h = g_{p} R_{p}$
$\frac{G M _{e}}{R _{e}^{2}} . h = \frac{G M _{p}}{R _{p}^{2}} R_{p}$
or $\frac{( \frac{4}{3} π R _{e}^{3} ) ρ h}{R _{e}^{2}} = \frac{( \frac{4}{3} π R _{p}^{3} ) ρ R _{p}}{R _{p}^{2}}$
or $R_{p} = R_{e} h$
   $= (6.41 \times 1 0^{6}) (1.5)$
   $= 3.1 \times 1 0^{3}$ m

Answer 2

h = \frac{u ^{2}}{2 g _{e}}

∴

u = 2 g_{e} h

(i)
For the asked planet this u sbould be equal to the escape velocity from its surface.

2 g_{e} h = 2 g_{p} R_{p}

or

g_{e} h = g_{p} R_{p}

\frac{G M _{e}}{R _{e}^{2}} . h = \frac{G M _{p}}{R _{p}^{2}} R_{p}

or

\frac{( \frac{4}{3} π R _{e}^{3} ) ρ h}{R _{e}^{2}} = \frac{( \frac{4}{3} π R _{p}^{3} ) ρ R _{p}}{R _{p}^{2}}

or

R_{p} = R_{e} h

= (6.41 \times 1 0^{6}) (1.5)

= 3.1 \times 1 0^{3}

m