Question

with what velocity must a body be thrown from earth's surface so that it may reach a height of 4Re above the earth's surface ?(Radius of the earth = 6400 km, g = 9.8 m/s2

Answer 1

Answer:

velocity = 10119 m/s

Explanation:

pls refer attachment

Answer 2

Dear Student,

◆ Answer -

v = 10015 m/s

● Explanation -

For body to reach certain height, kinetic energy equals to work done by the body.

Kinetic energy = Work done

1/2 mv^2 = GMm [(1/R - 1/(R+h)]

v^2 = 2GM × h / [R(R+h)]

Substitute h = 4R,

v^2 = 2GM × 4R / [R(R+4R)]

v^2 = 8GM/5R

As GM/R = gR,

v^2 = 8gR/5

v^2 = 8 × 9.8 × 6.4×10^6 / 5

v^2 = 1.003×10^8

v = 10015 m/s

Hence, body must be thrown at velocity of 10015 m/s.

Thanks dear. Hope this helps you...