Earth Satellites at a Glance

Earth satellites—natural (the Moon) or artificial—revolve around the planet in circular or elliptical paths, just like planets around the Sun. Because of that, the same three Kepler laws apply :contentReference[oaicite:0]{index=0}. Since 1957, artificial satellites have become key tools for telecommunication, geophysics, and weather monitoring :contentReference[oaicite:1]{index=1}.

Circular-Orbit Mechanics

  • Balancing forces. For a satellite of mass m in a circular orbit of radius RE + h:
    \( F_{\text{centripetal}}=\dfrac{m V^{2}}{R_{E}+h} \) (7.33)
    \( F_{\text{gravitation}}=\dfrac{G\,m\,M_{E}}{(R_{E}+h)^{2}} \) (7.34):contentReference[oaicite:2]{index=2}
  • Orbital speed. Equating the two forces gives \( V=\sqrt{\dfrac{G M_{E}}{R_{E}+h}} \) (7.35) —notice that the speed drops as altitude h rises :contentReference[oaicite:3]{index=3}.
  • Speed skimming the surface. Setting h = 0 yields \( V_{0} = \sqrt{g\,R_{E}} \) (7.36) :contentReference[oaicite:4]{index=4}.
  • Time for one lap. A satellite covers the circumference \(2\pi(R_{E}+h)\) at speed V, so \( T = \dfrac{2\pi(R_{E}+h)^{3/2}}{\sqrt{G M_{E}}} \) (7.37) :contentReference[oaicite:5]{index=5}.
  • Kepler’s period law. Squaring the last result gives \( T^{2}=k\,(R_{E}+h)^{3},\; k=\dfrac{4\pi^{2}}{G M_{E}} \) (7.38) :contentReference[oaicite:6]{index=6}.
  • Low-Earth “parking orbit.” When h is negligible, \( T_{0}=2\pi\sqrt{\dfrac{R_{E}}{g}}\approx85\;\text{min} \) (7.39) :contentReference[oaicite:7]{index=7}.

Energy Budget in Orbit

Using the speed from (7.35), the energies at radius RE+h are :contentReference[oaicite:8]{index=8}:

  • Kinetic: \( \text{KE} = \dfrac{G m M_{E}}{2(R_{E}+h)} \) (7.40)
  • Potential: \( \text{PE} = -\dfrac{G m M_{E}}{R_{E}+h} \) (7.41)
  • Total: \( E = \text{KE} + \text{PE} = -\dfrac{G m M_{E}}{2(R_{E}+h)} \) (7.42)

The negative total shows that a bound satellite can’t drift to infinity without extra energy. For elliptical orbits, replace RE+h by the semi-major axis; the total energy stays negative :contentReference[oaicite:9]{index=9}.

Worked Examples (Why They Matter!)

  • Mass of Mars from Phobos. Using Phobos’ period (7 h 39 min) and orbital radius (9.4 × 103 km) in Kepler’s law gives \( M_{m}\approx6.5\times10^{23}\,\text{kg} \) :contentReference[oaicite:10]{index=10}.
  • Weighing Earth two ways. (i) Directly from g and RE and (ii) from the Moon’s 27.3-day orbit; both yield \( M_{E}\approx6.0\times10^{24}\,\text{kg} \) with <1 % mismatch :contentReference[oaicite:11]{index=11}.
  • Unit conversion trick. Converting \(k=10^{-13}\,{\rm s^{2}m^{-3}}\) to d2 km-3 and checking the Moon again returns 27.3 days :contentReference[oaicite:12]{index=12}.

Quick-Fire NEET Essentials

  1. The orbital-speed formula \( V=\sqrt{\dfrac{G M_{E}}{R_{E}+h}} \) and how speed falls with altitude.
  2. Kepler’s third law for satellites \( T^{2}\propto(R_{E}+h)^{3} \) and its shortcut constant \(k\).
  3. Low-Earth time period \( T_{0}=2\pi\sqrt{\tfrac{R_{E}}{g}}\approx85\;\text{min}\) – a favorite objective-type question.
  4. Negative total orbital energy \( E=-\dfrac{G m M_{E}}{2(R_{E}+h)} \) – explains why satellites stay bound.
  5. “Weighing” planets via satellite periods (mass-from-orbit problems like Phobos and the Moon).

You now have the key ideas, formulas, and numerical tricks to tackle almost any satellite question with confidence—happy problem-solving!