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andrey2020 [161]
3 years ago
6

A man can jump 1.5 m on earth. calculate the approximate

Physics
1 answer:
Olegator [25]3 years ago
5 0

Answer:

18 m

Explanation:

G = Gravitational constant

m = Mass of planet = \rho V

\rho = Density of planet

V = Volume of planet assuming it is a sphere = \dfrac{4}{3}\pi r^3

r = Radius of planet

Acceleration due to gravity on a planet is given by

g=\dfrac{Gm}{r^2}\\\Rightarrow g=\dfrac{G\rho V}{r^2}\\\Rightarrow g=\dfrac{G\rho \dfrac{4}{3}\pi r^3}{r^2}\\\Rightarrow g=\dfrac{4G\rho\pi r}{3}

So,

g\propto \rho r

Density of other planet = \rho_p=\dfrac{1}{4}\rho_e

Radius of other planet = r_p=\dfrac{1}{3}r_e

\dfrac{g_e}{g_p}=\dfrac{\rho_e r_e}{\rho_p r_p}\\\Rightarrow \dfrac{g_e}{g_p}=\dfrac{\rho_e r_e}{\dfrac{1}{4}\rho_e\times \dfrac{1}{3}r_e}\\\Rightarrow \dfrac{g_e}{g_p}=12\\\Rightarrow g_p=\dfrac{g_e}{12}\\\Rightarrow g_p=\dfrac{9.8}{12}

Since the person is jumping up the acceleration due to gravity will be negative.

From kinematic equations we have

v^2-u^2=2g_es\\\Rightarrow u^2=v^2-2g_es\\\Rightarrow u^2=0-2\times -9.8\times 1.5\\\Rightarrow u^2=2\times 9.8\times 1.5

On the other planet

v^2-u^2=2g_ps\\\Rightarrow s=\dfrac{v^2-u^2}{2g_p}\\\Rightarrow s=\dfrac{0-(2\times 9.8\times 1.5)}{2\times -\dfrac{9.8}{12}}\\\Rightarrow s=18\ \text{m}

The man can jump a height of 18 m on the other planet.

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