Question

If an object is projected vertically upwards with speed, half the escape speed of earth, then the maximum height attained by it is (R is radius of the earth)..?

Answer 1

According to the law of conservation of energy,

(T.E)at surface = (T.E) at height

(K.E+P.E) surface = (K.E+P.E) max height(

1/2mv2 + (−GMm/R) =0+(−GMm/R+h)

Given
V=3/4ve =3/4√2GM/R

=1/2m × 9/16 × (2GM/R) + (−GMm/R)

=(−GMm/R+h)

9GMm/16R − GMm/R = −GMm/R+h

−7GMm/16R = −GMm/R+h7/16R = 1/R+h⇒7R+7h=16Rh=9/7R

Answer 2

Explanation:

We know that the escape velocity of Earth = Ve = √[2gR]

       g = gravity due to Earth

      1/2 m (Ve)² - G Mm/R = 0   for the object to escape from Earth.

       (or,  KE + PE = 0)

When the body is projected from Earth's surface:

     Given  u = initial velocity = Ve /2 = √(gR/2)

     Initial  KE = 1/2 m u² = m g R/4 = G M m /(4R)

     Initial  PE = - GMm/R

     Total initial energy = - 3 GMm /(4R)

 When the body reaches a height h above surface of Earth and stops:

     v = 0. Final KE = 0.

     Final PE = - G M m/ (R+h)

From conservation of Energy:

           - GMm/(R+h) = - 3Gm/ (4R)

            3 (R+h) = 4 R

              h = R/3

Answer is R/3.