A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be :
(Take radius of earth =6400 km and g=10 ms⁻² )

Solution

Physics sir · 4 steps

Use energy conservation with -GMm/r, not mgh, because g does not stay constant up to such heights.

1. Launch speed. Escape velocity is vₑ = √2gR, so v = vₑ/3 gives v² = 2gR/9. Also write GM = gR² so everything is in terms of g and R.
2. Conserve energy. At the surface the body has kinetic energy plus potential energy -GMm/R. At the top it has zero kinetic energy and potential energy -GMm/(R+h).

   1/2m(2gR/9) - mgR²/R = -mgR²/(R+h)

3. Solve for h. Cancel mgR to get 1/9 - 1 = -R/(R+h), so R/(R+h) = 8/9. That gives 9R = 8R + 8h, so h = R/8.
4. So. h = 6400/8 = 800 km, which is option (a).

The idea to keep. For a launch at speed v, the height reached is h = R/((vₑ/v)² - 1). Square the speed ratio before you subtract 1.

Answer from the NTA final answer key. Solution by PrepWiser.