Question

The required equation to solve this problem :- ( v^{2}=u^{2}-2 g h . v ) is final speed in this case it is zero, when astronaut jumps and reaches maximum height his speed is zero. u is initial jumping speed. g is acceleration due to gravity and h is the maximum jumping height.
If jumping speed on earth and moon are assumed as same, then height is inversely proportional to acceleration due to gravity.
( h propto frac{1}{g} )
now acceleration due to gravity ( g=frac{C M}{R^{2}}=frac{C}{R^{2}} frac{4}{3} pi R^{3} rho=C frac{4}{3} pi R rho )
( mathrm{G} ) - gravitational constant M mass of earth, ( rho ) is mean density, ( mathrm{R} ) is radius
If we consider ( 2 / 3 ) of earth density as moon's density and ( 1 / 2 ) of earth radius as moon's radius then acceleration due to gravity on moon, ( g_{M}=frac{1}{6} g_{E} )
( mathrm{g} E ) is acceleration due to gravity on earth
( frac{text {jumping height on Moon}}{text {jumping height on Earth}}=frac{g_{E}}{g_{M}}=frac{g_{E}}{(1 / 6) g_{E}}=6 )
Hence the astronaut will jump a height of ( 6 times 0.5=3 mathrm{m} ) on moon

# The maximum vertical distance through which a full dressed astronaut can jump on the earth is 0.5 m. Estimate the maximum vertical distance through which he can jump on the moon, which has a mean density 2/3rd that of the earth and radius one quarter that of the earth.

Solution