Escape Speed

Think of escape speed as the slowest “run-up” an object needs so Earth’s gravity can never tug it back. In other words, once the object climbs out to infinity, it still has just enough energy to keep going (its final speed there can be zero).:contentReference[oaicite:0]{index=0}

Energy Story

Total mechanical energy E is the sum of kinetic and gravitational potential energies:

  • At infinity: \(E_\infty = \tfrac12\,mV_f^{2}\). Setting the “zero” of potential energy at infinity makes the math cleaner.:contentReference[oaicite:1]{index=1}
  • At launch point (distance \(R_E+h\) from Earth’s centre): \(E_{\text{launch}} = \tfrac12\,mV_i^{2}\;-\;\dfrac{G\,M_E\,m}{R_E+h}\).:contentReference[oaicite:2]{index=2}

Energy conservation says \(E_{\text{launch}} = E_\infty\). With the smallest possible \(V_f = 0\), you arrive at:

\[ V_i \;=\;\sqrt{\frac{2\,G\,M_E}{R_E+h}} \]:contentReference[oaicite:3]{index=3}

The Special Case: Launching from Earth’s Surface

Put \(h = 0\) in the general formula and swap \(\dfrac{G\,M_E}{R_E}\) with \(gR_E\):

\[ V_i\;=\;\sqrt{2\,g\,R_E}\;\approx\;11.2\;\text{km s}^{-1} \]:contentReference[oaicite:4]{index=4}

On the Moon the same square-root rule, with lunar \(g\) and \(R\), gives only \(2.3\;\text{km s}^{-1}\). That tiny value lets gas molecules drift away, which is why the Moon has no significant atmosphere.:contentReference[oaicite:5]{index=5}

Worked Example: Two-Sphere Challenge

Imagine two solid spheres of radius \(R\) separated by \(6R\). The lighter one has mass \(M\); the heavier has \(4M\). A projectile starts on the lighter sphere and heads straight toward the heavier.:contentReference[oaicite:6]{index=6}

  1. The gravitational pulls cancel at the “neutral point” \(2R\) from the lighter sphere.
  2. Give the projectile just enough speed to coast to that point; afterwards the heavier sphere does the rest.
  3. The minimum launch speed works out to \(v = \sqrt{\dfrac{3\,G\,M}{5\,R}}\).:contentReference[oaicite:7]{index=7}

Handy Background Facts

  • Gravitational potential energy between two masses \(m_1\) and \(m_2\): \(U = -\dfrac{G\,m_1\,m_2}{r}\).:contentReference[oaicite:8]{index=8}
  • Choosing \(U = 0\) at infinity lets you talk only about meaningful energy differences.:contentReference[oaicite:9]{index=9}

High-Yield Ideas for NEET

  1. Escape-speed formula \(V_e = \sqrt{2gR_E}\;(\;11.2\;\text{km s}^{-1}\;)\).:contentReference[oaicite:10]{index=10}
  2. Planetary dependence Lower \(g\) or smaller \(R\) ⇒ lower \(V_e\); that’s why moons and small planets lose atmospheres.:contentReference[oaicite:11]{index=11}
  3. Energy-conservation method Equate kinetic plus potential energy at launch with energy at infinity to derive escape conditions fast.:contentReference[oaicite:12]{index=12}
  4. Neutral point in multi-body fields Know how to locate where gravitational pulls cancel and how to find the projectile’s minimum speed.:contentReference[oaicite:13]{index=13}
  5. Core potential-energy law \(U = -\dfrac{G\,m_1\,m_2}{r}\) underpins every gravitation energy calculation.:contentReference[oaicite:14]{index=14}