Question

A body falls through a distance H in a certain time on earth . Then if the same body is related on another planet having mass and radius twice as that of earth , the distance through which it falls in the same time is ?​

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

Read more on Brainly.in - https://brainly.in/question/7687823#readmoreStep-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

Read more on Brainly.in - https://brainly.in/question/7687823#readmoreStep-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

Read more on Brainly.in - https://brainly.in/question/7687823#readmore

Explanation: