Satellite motion — orbital velocity, escape velocity, binding energy comparison

Question

Derive the relationship between orbital velocity, escape velocity, and binding energy of a satellite. If the orbital velocity of a satellite near Earth’s surface is $v_o$ , what is its escape velocity?

(CBSE 11 + JEE Main + NEET)

Solution — Step by Step

For a satellite in circular orbit, gravity provides the centripetal force:

\frac{GMm}{r^2} = \frac{mv_o^2}{r}

v_o = \sqrt{\frac{GM}{r}}

Near Earth’s surface ( $r \approx R$ ): $v_o = \sqrt{gR} \approx \sqrt{9.8 \times 6400 \times 10^3} \approx \mathbf{7.9 \text{ km/s}}$

Escape velocity is the speed needed to reach infinity with zero final velocity. Using energy conservation:

\frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0

v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} \approx \mathbf{11.2 \text{ km/s}}

v_e = \sqrt{2} \cdot v_o

This means a satellite already in orbit needs only a $(\sqrt{2} - 1) \approx 41\%$ boost in speed to escape Earth’s gravity. This is why rockets first achieve orbit, then fire again to escape — it is more fuel-efficient.

Binding energy = energy needed to remove the satellite from orbit to infinity.

Total energy in orbit: $E = KE + PE = \frac{1}{2}mv_o^2 - \frac{GMm}{r} = -\frac{GMm}{2r}$

\text{Binding Energy} = |E| = \frac{GMm}{2r} = \frac{1}{2}mv_o^2

The satellite’s binding energy equals its kinetic energy. To escape, we need to supply this much energy.

flowchart TD
    A["Satellite Parameters"] --> B["Orbital velocity: v_o = √(GM/r)"]
    A --> C["Escape velocity: v_e = √(2GM/r)"]
    A --> D["Binding Energy = GMm/(2r)"]
    B --> E["v_e = √2 × v_o"]
    C --> E
    D --> F["BE = KE in orbit = ½mv_o²"]
    A --> G{"What changes with altitude?"}
    G --> H["Higher orbit → slower v_o"]
    G --> I["Higher orbit → longer time period"]
    G --> J["Higher orbit → less binding energy"]

Why This Works

A satellite in orbit is in free fall — gravity continuously pulls it inward, but its tangential velocity keeps it from falling. The balance between gravitational pull and the centripetal requirement fixes the orbital speed.

Escape velocity comes from energy conservation: the satellite needs just enough kinetic energy to overcome the gravitational potential well. The factor of $\sqrt{2}$ between $v_e$ and $v_o$ is elegant — it means an orbiting satellite is already more than halfway (in energy terms) to escaping.

Alternative Method

For quick calculations, use $v_o \approx 8$ km/s and $v_e \approx 11.2$ km/s near Earth’s surface. For a satellite at height $h$ , replace $R$ with $R + h$ in all formulas. The time period is $T = 2\pi r/v_o = 2\pi\sqrt{r^3/(GM)}$ (Kepler’s third law). Near the surface, $T \approx 84$ minutes.

Common Mistake

Students forget that escape velocity does NOT depend on the mass of the escaping object — it depends only on the mass and radius of the body being escaped from. A 1 kg satellite and a 1000 kg satellite both need the same escape velocity (11.2 km/s from Earth). The energy required differs ( $\frac{1}{2}mv_e^2$ ), but the speed is the same. This appears as a trick MCQ in both JEE and NEET.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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