MHT-CET Physics · Gravitation

Gravitational Potential Energy and Escape Velocity

A mass m at distance r from the centre of a mass M has potential energy −GMm/r; conserving kinetic plus potential energy between two distances gives a falling body's speed, a projected body's greatest height, and the escape velocity √(2GM/R) at which it never returns.

All Gravitation notes MHT-CET Physics strategy

Why this matters

19 PYQs, 8 of them HARD — the most HARD questions in the chapter. Nine use potential energy between two distances: the speed a body gains falling from far away, the energy to raise a satellite, a satellite's energy ratios. Ten are escape velocity and the height reached below it, including escape from a point between the earth and the moon. Two cards.

Concept 1 of 2: Energy Conservation Between Two Distances

Gravitational potential energy −GMm/r is negative and rises toward zero with distance. Energy conservation between r₁ and r₂ gives ½mv² = GMm(1/r₂ − 1/r₁) for a body falling inward; a body released at R₀ reaches the surface at √(2GM(1/R − 1/R₀)). Measure every distance from the CENTRE: a body '3R above the surface' is 4R out. The energy to lift a body to height h is GMm h/(R(R + h)) = mgh/(1 + h/R); setting it in orbit there needs the orbital kinetic energy GMm/2(R + h) as well, so the two are in the ratio 2h : R. In a circular orbit the potential energy is −2 times the kinetic energy.

Definition

$U = -\dfrac{GMm}{r}$ ; $\tfrac{1}{2}mv^2 = GMm\left(\dfrac{1}{r_2} - \dfrac{1}{r_1}\right)$ .
From 3R above the surface to R above: $KE = \dfrac{GMm}{2R} - \dfrac{GMm}{4R} = \dfrac{GMm}{4R}$ .
Magnitude E at distance R ⇒ weight at 1.5R $= \dfrac{4E}{9R}$ .
Raising to h: $E_1 = \dfrac{mgh}{1 + h/R}$ ; orbital KE there: $E_2 = \dfrac{mgR}{2(1 + h/R)}$ ; $E_1 : E_2 = 2h : R$ .
Projectiles near the ground: heights and so PE at the top go as the square of the vertical velocity component (vertical vs 60° to the vertical ⇒ 4 : 1).

Gravitational potential energy

U = -\frac{GMm}{r}, \qquad \tfrac{1}{2}mv^2 = GMm\left(\frac{1}{r_2} - \frac{1}{r_1}\right)

Worked example

A body is released from rest at 2R from the earth's centre. Its speed on reaching the surface, in terms of g and R?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 1 · Q35Hard

Example 1 · Gravitation · Gravitational PE, Escape Velocity, and Energy

A body of mass m kg starts falling from a distance 3R above earth's surface. When it reaches a distance R above the surface of the earth of radius R and mass M, then its kinetic energy is

Measuring from the surface instead of the centre

Potential energy uses the distance from the centre. '3R above the surface' is 4R from the centre, and the answers differ by a factor of two.

Using mgh far from the earth

mgh assumes g does not change. For heights comparable to R, use −GMm/r at both ends.

Drill 8 more on energy conservation between two distances

Concept 2 of 2: Escape Velocity and the Height Below It

To escape, a body needs enough kinetic energy to bring its total energy to zero: ½mv_e² = GMm/R, so v_e = √(2GM/R) = √(2gR), about 11.2 km/s on earth. It does not depend on the body's mass. For planets of the same density v_e ∝ R, and in general v_e ∝ R√ρ. Projected at a fraction n of v_e, the body rises to h = n²R/(1 − n²): a third of v_e reaches R/8, half reaches R/3. Between two bodies, the escape speed comes from the total potential energy at the launch point.

Definition

$v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR} = R\sqrt{\dfrac{8\pi G\rho}{3}}$ : independent of the body's mass; $v_e \propto R\sqrt{\rho}$ .
Radius 2R, same density ⇒ 22 km/s; radius and density both ×4 ⇒ ×8.
Projected at $nv_e$ : $h = \dfrac{n^2R}{1 - n^2}$ ( $\tfrac{1}{3}v_e$ ⇒ R/8; $\tfrac{1}{2}v_e$ ⇒ R/3).
Midway between masses M₁, M₂ a distance d apart: $v = 2\sqrt{\dfrac{G(M_1 + M_2)}{d}}$ .

Escape velocity

v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}, \qquad h = \frac{n^2R}{1 - n^2}

Worked example

A body is projected upward at 2/3 of the escape velocity. How high does it rise?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q28Hard

Example 2 · Gravitation · Gravitational PE, Escape Velocity, and Energy

A body is projected in vertically upward direction from the surface of the earth of radius 'R' into space with velocity 'nVe'

(n<1)

. The maximum height from the surface of earth to which a body can reach is

Using h = v²/2g for a large launch speed

At a third of the escape velocity, v²/2g gives R/9; the true height, with g falling off, is R/8.

Scaling escape velocity with √R at fixed density

v_e = R√(8πGρ/3) is proportional to R when the density is fixed, so twice the radius doubles it.

Drill 9 more on escape velocity and the height below it

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Energy Conservation Between Two Distances
Gravitational potential energy
$U = -\frac{GMm}{r}, \qquad \tfrac{1}{2}mv^2 = GMm\left(\frac{1}{r_2} - \frac{1}{r_1}\right)$
Escape Velocity and the Height Below It
Escape velocity
$v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}, \qquad h = \frac{n^2R}{1 - n^2}$

Watch out for (4)

Measuring from the surface instead of the centre
→ Energy Conservation Between Two Distances
Using mgh far from the earth
→ Energy Conservation Between Two Distances
Using h = v²/2g for a large launch speed
→ Escape Velocity and the Height Below It
Scaling escape velocity with √R at fixed density
→ Escape Velocity and the Height Below It

Test yourself on Gravitation

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Gravitation chapter test Drill all 19 questions