Use Gravitational Potential Energy:
Gravitational PE at distance r from Earth's centre:
$$U = -\frac{GMm}{r}$$
At surface (r = R): \(U_i = -\dfrac{GMm}{R}\)
At height h = R (r = 2R): \(U_f = -\dfrac{GMm}{2R}\)
Work done = Change in PE:
$$W = U_f - U_i = -\frac{GMm}{2R} - \left(-\frac{GMm}{R}\right)$$
$$W = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$$
Substitute GM = gR²:
Since \(g = \dfrac{GM}{R^2}\), we have \(GM = gR^2\)
$$W = \frac{gR^2 \cdot m}{2R} = \mathbf{\frac{mgR}{2}}$$
Quick Formula Method:
$$W = \frac{mgh}{1 + h/R} = \frac{mgR}{1 + 1} = \frac{mgR}{2}$$ ✓
Note: Never use W = mgh for large heights since g is NOT constant.
Every two masses in the universe attract each other with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them:
\(F = \dfrac{GMm}{r^2}\)
G = 6.674 × 10⁻¹¹ N m²/kg² is the universal gravitational constant. This force is always attractive — there is no repulsive gravity. It acts along the line joining the centres of the two masses and obeys Newton's third law.
Gravitational potential energy is defined as the work done to bring a mass from infinity to a given point in the gravitational field. Since gravity is attractive, work is done by gravity (not against it) when bringing a mass from infinity — hence the negative sign:
\(U = -\dfrac{GMm}{r}\)
U = 0 at r = ∞ (reference point). U is negative everywhere (bound state). As r decreases (object moves closer to Earth), U becomes more negative — the object loses potential energy and the gravitational field does positive work.
The formula W = mgh assumes g is constant throughout the journey. This is valid only when h << R (say, h less than a few kilometres). For large heights like h = R, g decreases significantly with altitude. At h = R, g is only g/4 of its surface value. Using W = mgh = mgR would give double the correct answer!
⚠️ W = mgh: valid only when h << R
⚠️ W = mgR/2: correct answer for h = R
⚠️ Always use W = mgh/(1 + h/R) for large heights
\(W = \dfrac{mgh}{1 + h/R} = \dfrac{mgRh}{R + h}\)
📌 h = R: W = mgR/2
📌 h = 2R: W = 2mgR/3
📌 h = 3R: W = 3mgR/4
📌 h → ∞: W = mgR (escape energy from surface)
📌 h << R: W ≈ mgh (approximate, near surface)
Above surface (height h): \(g_h = g\left(\dfrac{R}{R+h}\right)^2 = \dfrac{g}{(1+h/R)^2}\). At h = R: g_h = g/4.
Below surface (depth d): \(g_d = g\left(1 - \dfrac{d}{R}\right)\). At centre (d = R): g = 0. The variation is linear inside Earth and follows inverse square law outside.
📌 g decreases with both height (above) and depth (below)
📌 g is maximum at the Earth's surface
📌 g = 0 at Earth's centre
📌 g variation with height: 1/r² law (outside)
📌 g variation with depth: linear (inside)
Escape velocity is the minimum initial velocity needed to project an object from Earth's surface so that it escapes the gravitational field (reaches r = ∞ with zero KE). Setting total mechanical energy = 0:
\(v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR} \approx 11.2 \text{ km/s}\)
Note that escape velocity is independent of the mass of the projected object. It depends only on the planet's mass and radius. The Moon's escape velocity is only ~2.4 km/s (weaker gravity), which is why the Moon has no atmosphere — gas molecules can escape easily.
For a satellite in circular orbit at height h (orbit radius r = R + h), gravitational force provides centripetal force:
\(v_o = \sqrt{\dfrac{GM}{R+h}} = \sqrt{\dfrac{gR^2}{R+h}}\)
For near-Earth orbit (h << R): v_o ≈ √(gR) ≈ 7.9 km/s. Note: v_e = √2 × v_o, so escape velocity is always √2 times the orbital velocity at the same radius.