Skip to content
III. Long Answer Questions · Q12

Q.Derive an expression for energy of a satellite.

Puducherry TnboardTextbookSubjectiveImportance★★★★★
63% · 55/88 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Step 1. The gravitational potential energy of a satellite of mass MsM_s orbiting at height hh is U=−GMsMeRe+hU=-\dfrac{GM_sM_e}{R_e+h}.

Step 2. Its kinetic energy, using the orbital speed v=GMe/(Re+h)v=\sqrt{GM_e/(R_e+h)} from the circular-orbit condition, is

K.E.=12Msv2=12⋅GMsMeRe+h.K.E.=\frac12M_sv^2=\frac12\cdot\frac{GM_sM_e}{R_e+h}.

Step 3. Adding the two:

E=K.E.+U=12GMsMeRe+h−GMsMeRe+h=−12GMsMeRe+h.E=K.E.+U=\frac12\frac{GM_sM_e}{R_e+h}-\frac{GM_sM_e}{R_e+h}=-\frac12\frac{GM_sM_e}{R_e+h}.

Step 4. The negative sign means the satellite's total mechanical energy is less than the energy it would have if completely free of Earth's gravity at rest at infinity (where E=0E=0) -- so the satellite is permanently gravitationally bound to the Earth. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.