Energy of an Orbiting Satellite

1 · Kinetic Energy (K)

A satellite of mass m moving in a circular orbit of radius \(R_E+h\) has

$$K=\frac12\,m\,v^{2} =\frac{G\,M_E\,m}{2\,(R_E+h)}$$ :contentReference[oaicite:1]{index=1}

2 · Gravitational Potential Energy (U)

Taking the zero of potential energy at infinity, the value at distance \(R_E+h\) is

$$U=-\frac{G\,M_E\,m}{(R_E+h)}$$ :contentReference[oaicite:3]{index=3}

3 · Total Mechanical Energy (E)

The satellite’s total energy combines K and U:

$$E=K+U=-\frac{G\,M_E\,m}{2\,(R_E+h)}$$ :contentReference[oaicite:5]{index=5}

You can spot two handy relations:

  • \(K=-\tfrac12\,U\)
  • \(U=2\,E\)

Take-away: the total energy is negative because the satellite is bound to Earth, and the potential energy’s magnitude is twice that of the kinetic energy (but opposite in sign). :contentReference[oaicite:7]{index=7}

4 · Elliptical Orbits

In an ellipse the values of K and U change from point to point, yet the sum E stays constant and negative. A satellite with E that is zero or positive would escape Earth’s pull, so bound satellites always have negative total energy. :contentReference[oaicite:9]{index=9}

5 · Worked Example (Example 7.8)

Problem —A 400 kg satellite orbits at \(2R_E\). How much extra energy pushes it to a new circular orbit at \(4R_E\)? What happens to its kinetic and potential energy?

  • Initial energy: \(E_i=-\dfrac{G\,M_E\,m}{4R_E}\)
  • Final energy:   \(E_f=-\dfrac{G\,M_E\,m}{8R_E}\)
  • Energy required (work done on the satellite): $$\Delta E=E_f-E_i\approx 3.13\times10^{9}\,\text{J}$$
  • Kinetic-energy change: $$\Delta K=K_f-K_i\approx-3.13\times10^{9}\,\text{J}$$
  • Potential-energy change: $$\Delta U=U_f-U_i\approx-6.25\times10^{9}\,\text{J}$$

The transfer needs about \(3.1\times10^{9}\) joules of extra energy, while the satellite slows down in its higher orbit. :contentReference[oaicite:11]{index=11}

6 · Quick Recap

  • Bound orbits have negative total energy: E < 0.
  • For a circular path: \(K=-\tfrac12\,U\) and \(E=-\tfrac12\,U\).
  • Moving to a higher orbit (larger radius) demands energy input even though the satellite’s speed drops.
  • In an ellipse, E still equals \(-\dfrac{G\,M_E\,m}{2a}\), with a the semi-major axis (replace \(R_E+h\) by a). :contentReference[oaicite:13]{index=13}

High-Yield Ideas for NEET

  1. Energy trio: memorize \(K=\dfrac{G\,M_E\,m}{2\,(R_E+h)}\), \(U=-\dfrac{G\,M_E\,m}{(R_E+h)}\), and \(E=-\dfrac{G\,M_E\,m}{2\,(R_E+h)}\). These pop up often. :contentReference[oaicite:15]{index=15}
  2. Sign check: total energy of a satellite is always negative; \(E\ge0\) means escape. :contentReference[oaicite:17]{index=17}
  3. Relation \(K=-\tfrac12 U\): a quick way to switch between K and U in calculations. :contentReference[oaicite:19]{index=19}
  4. Elliptical twist: replace \(R_E+h\) with the semi-major axis a to handle non-circular orbits. :contentReference[oaicite:21]{index=21}
  5. Orbit transfer energy: practice Example 7.8 style questions—energy change equals the difference in total energies of the two orbits. :contentReference[oaicite:23]{index=23}