Gravitational Potential Energy
Gravitational potential energy tells you how much energy a mass stores because of where it sits in Earth’s gravity field. Move the mass, and you change that energy. The sections below show you how to calculate the change in three everyday situations.
1. How g changes when you go below the surface
Only the sphere of radius \(R_E – d\) pulls on a mass at depth \(d\). That gives a force
$$F(d)=\frac{G M_E m\,(R_E-d)}{R_E^{3}}$$
and an acceleration
$$g(d)=g\!\left(1-\frac{d}{R_E}\right).$$
Take-away: \(g\) shrinks linearly as you dig deeper.:contentReference[oaicite:0]{index=0}
2. Near-surface potential energy (constant g)
Close to the ground \(g\) hardly changes, so lifting a mass \(m\) from height \(h_1\) to \(h_2\) needs work
$$W_{12}=mg(h_2 – h_1).$$
If you call the ground your zero, the energy at height \(h\) is simply
$$W(h)=mgh.$$:contentReference[oaicite:1]{index=1}
3. Potential energy at any distance \(r\) from Earth’s centre
Farther out you still feel a force
$$F(r)=\frac{G M_E m}{r^{2}}.$$
Doing the work from \(r_1\) to \(r_2\) gives
$$W_{12}=G M_E m\!\left(\frac{1}{r_1}-\frac{1}{r_2}\right).$$
If you set the zero of energy at infinity, a mass at distance \(r\) stores
$$W(r)=-\frac{G M_E m}{r}.$$:contentReference[oaicite:2]{index=2}
4. Gravitational potential V
Divide the energy by the test mass and you find
$$V(r)=-\frac{G M_E}{r}.$$
For two masses \(m_1\) and \(m_2\) separated by \(r\)
$$U=-\frac{G m_1 m_2}{r}.$$
Add the pair-wise terms to get the energy of any group of masses (superposition idea).:contentReference[oaicite:3]{index=3}
5. Worked example: four equal masses on a square
Place four masses \(m\) at the corners of a square of side \(l\). Four edges contribute \(-\displaystyle\frac{G m^{2}}{l}\) each, and two diagonals add \(-\displaystyle\frac{G m^{2}}{\sqrt{2}l}\) each. The total works out to
$$U=-\frac{5.41\,G m^{2}}{l}.$$
The potential at the square’s centre is
$$V=-\frac{4\sqrt{2}\,G m}{l}.$$:contentReference[oaicite:4]{index=4}
6. Escape speed—why you need a big launch
Imagine flinging a rocket so hard that it slows to a stop only when it reaches infinity. At infinity the mechanical energy is zero. Set the same energy at the launch point (distance \(R_E + h\) from Earth’s centre) and you get
$$-\frac{G M_E m}{R_E+h}+\frac{1}{2}mV_i^{2}=0.$$
Solving for \(V_i\) gives the escape speed—the minimum speed that lets the rocket break free.:contentReference[oaicite:5]{index=5}
High-Yield Ideas for NEET
- Linear fall-off of \(g\) with depth: \(g(d)=g\!\left(1-\dfrac{d}{R_E}\right).\)
- Near-surface energy change: \(W=mgh\) with zero at ground level.
- General potential energy: \(W(r)=-\dfrac{G M_E m}{r}\) (zero at infinity).
- Energy for two bodies: \(U=-\dfrac{G m_1 m_2}{r}\) and superposition for many bodies.
- Escape condition: initial kinetic energy must equal \(\dfrac{G M_E m}{R_E+h}.\)
