A particle of mass ‘m’ is projected with a velocity v=kV_e (k < 1) from the surface of the earth. (V_e = escape velocity) The maximum height above the surface reached by the particle is :

**Given:** A particle of mass  is launched from the Earth’s surface with speed v = k Vₑ, where k < 1 and Vₑ is the escape velocity.

**Step 1 – Escape velocity:** The escape velocity from a planet of mass M and radius R is

Thus Vₑ² = 2GM/R.

**Step 2 – Initial kinetic energy:** The particle’s initial speed is v = k Vₑ, so

**Step 3 – Initial gravitational potential energy:** At the Earth's surface the potential energy is

**Step 4 – Total mechanical energy at launch:**

Since k < 1, this energy is negative – the particle is bound to the Earth.

**Step 5 – Energy at the highest point:** At the maximum height the speed becomes zero, so kinetic energy vanishes and only gravitational potential remains:

where r = R + h is the distance from the Earth's centre.

**Step 6 – Equate the two expressions for E:**

Cancel GMm from both sides:

Multiply by –1:

Solve for r:

**Step 7 – Height above the surface:** = r − R = R [ 1/(1 − k²) − 1 ] = R · k² / (1 − k²).

Therefore the maximum height attained by the particle is

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