Question

A projectile is launched from the surface of the earth (RE-6.37x10 m, ME-6.0x10 kg) with an initial speed of 10,000 m/s. Ignoring air resistance (or pretending air resistance was offset by additional rocket thrust)

a) what is the object's maximum distance from the center of the earth? Or, if it does not fall back to earth, what is its speed when far away?

B)what is the object's speed as enter space (conventionally defined as a height of 100km above the surface)

Answer 1

This problem can be solved by energy conservation.

We know potential energy (P.E.) of object of mass m kept on earth surface is U= - GMm/R

kinetic energy (K.E.) of object = mv²/2 ( where v is initial velocity with which particle is projected)

At maximum height H, K.E. is zero & P.E . is = - GMm / (R+H)

Now using energy conservation, we get

Initial Energy = Final Energy

-GMm / R + mv² / 2 = -GMm / (R+H)

on simplifying above we get,

R + H = R² / ( R- v² / 2g )

= (6370)² / (6370- (10⁴)² / 2 X10) ( here I take g=10 m/s²)

= 29618.17 Km

i.e. object is  29618.17 Km far from center of earth.

and H= (29618.17 - 6370) Km = 23248.17 Km far from earth surface.

Initial velocity required to project an object from earth surface to far away (say, infinite) is,

. This is well known escape velocity on earth surface.   

11.2 X 10³ m s^-1 11.2 Km s^-1

Object will maintain this speed upto infinite height from earth surface.

(b) Solution :

We know velocity of projection to project an object from earth surface to height h is given by

Therefor object's speed at Height 100Km  above the surface is,