Question

a body Falls through a distance H in a certain time on the earth then if the same body is released on another planet having mass and radius twice as that of the earth the distance through which it falls in the same time will be

Answer 1

Answer:

Step-by-step explanation:

S = ut + 1/2 gt^2 (second law of motion)

H= 1/2gt^2 ( u=0 ) ............{1}

t =√2H/g (solving equation 1).............{2}

Let mass of earth be M and radius be R

Therefore according to the question radius of planet =2R and mass =2M

Acceleration due to gravity on planet =

G×mass of planet ÷ (radius of planet)^2

here G=gravitational constant not acceleration due to garvity

Putting values we get

g=G2M÷4R^2

Second law of motion

Height at which ball is thrown from planet=1/2gt^2

= 1/2 ×(G×2M)÷4R^2×2H÷g

Hence H =GMH÷R^2g

Answer 2

Answer:

Step-by-step explanation:

If the body is dropped in earth's gravitational field, d is the distance travelled in time t , then we have

 

d = (1/2) g t2 ...............(1)

 

where g is acceleration due to gravity in earth's gravitational field .

 

If the body is dropped in some other planet's gravitational field, d' is the distance travelled in time t , then we have

 

d' = (1/2) g' t2 ................(2)

 

where g' is acceleration due to gravity in other planet's gravitational field

 

By dividing eqn.(2) by eqn.(1), we get,  d'/d = g'/g  ........................(3)

 

we have, acceleration due to gravity on earth,  g  = GM/R2 .......................(4)

 

where G is universal gravitation constant, M is mass of earth and R is radius of earth

 

we have, acceleration due to gravity on other planet,  g'  = G(2M)/(2R)2 = (1/2) GM/R2 .......................(5)

 

Hence, from eqn.(4) and (5), we get,  g'/g = (1/2) .....................(6)

 

using eqns.(3) and (6), we get d'/d = (1/2)   or  d' = d/2