Question

An object falls a distance H in 50 s when dropped on the surface of the earth. How
long would it take for the same object to fall through the same distance on the
surface of a planet whose mass and radius are twice that of the earth ? (Neglect air
resistance.)

Answer 1

Answer:

नोनू जिंदाबाद प्रो प्रो मदरबोर्ड

Answer 2

g1 = 10m/s^2 (As object is thrown on earth)
u = 0m/s^2
t = 50m/s^2

s = ut + 1/2 at^2
s = ut + 1/2 g1*t^2 ( a = g1 )
s = 1/2 * 10 * 50^2
s = 5 * 2500
s = 12500 m
Therefore, object is falling from 12500 m

Mass of other planet (m2) = 2m1 ( 2 * mass of earth )
Radius of other planet (r2) = 2r1 ( 2 * radius of earth )

g2 = Gm2 / (r2)^2
=G*2*m1/(2*r1)^2
=2*Gm1/4*r1
= Gm1/2*r1 ( g1 = Gm1/r1 )
= g1/2
= 10/2 ( g1 = 10 m/s^2 )
= 5 m/s^2

s = ut + 1/2 at^2
s = ut + 1/2 g2 * t^2
s = 1/2 g2 * t^2
12500 = 1/2 * 5 * t^2
25000/5 = t^2
t^2 = 5000
t = 70.7 s

Therefore , 70.7 s object will be required to fall on surface of planet which have twice the mass and radius of earth .

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