the maximum vertical distance through which a fully dressed astronaut can jump on the earth is 0.5 m. if mean density of the moon is 2/3rd that of the earth and radius is 1/4 that of earth,then what is the maximum vertical distance through which he can jump on the moon and the ratio of the time of duration of the jump on the moon to that of the earth ?

Suppose,Mass of earth is M, radius of earth is R.

Density of earth is, d = M/[(4/3)πR3]

Density of moon is, d/ = (2/3)d = (2/3) × M/[(4/3)πR3] = M/(2πR3)

Radius of moon is, R/ = R/4

Mass of moon is, M/ = (d/)[(4/3)π(R/)3] = [M/(2πR3)] × [(4/3)πR3/64] = M/96

Now,

An astronaut of mass ‘m’ jumps to a height 0.5 m on earth. Suppose he projects himself with initial velocity ‘u’

So, initial kinetic energy = potential energy at the height

=> ½ mu2 = mg(0.5)

=> u2 = (2)(0.5)g = g

=> u = (g)1/2

Using, v = u + at

=> -u = u + gt

=> t = 2u/g

This is the time of flight or duration of jump on earth.

On moon also the initial velocity is same.

So,

½ mu2 = m(g/)h

=> u2 = 2(g/)h

=> g = 2(g/)h

=> h = g/[2g/]

Now,

g = GM/R2

g/ = G(M/)/(R/)2 = G(M/96)/(R/4)2

=> g/ = (16/96)g = g/6

Therefore,

h = g/(2×g/6) = 3 m

Thus, on moon the astronaut jumps upto 3 m height.

Now, on moon the time of flight is found as,

v = u + at

=> -u = u + (g/)(t/)

=> t/ = 2u/g/

=> t/ = 2u/(g/6) = 6 × 2u/g

=> t/ = 6t

So, the ration of duration of jump on moon to that of on earth is, t / : t = 6 : 1

 

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