Gravitation — Part 2: Potential energy, escape speed and satellites: visual lesson, CBSE Class 11 Physics

Lesson

Energy in the gravitational field

Part 2 — potential energy, escape and satellites.

In this lesson you will

Define and calculate gravitational potential energy and potential
Derive and apply the escape speed
Derive the orbital speed and period of a satellite and its energy

Force towards the centre, velocity along the tangent.

Concept

Gravitational potential energy

Take the potential energy to be zero when the two masses are infinitely far apart. Bringing $m$ from infinity to distance $r$ from $M$ , gravity does positive work, so the energy becomes negative:

As $m$ moves in from $x = \infty$ to $x = r$ gravity does work $W = \int_r^{\infty} \dfrac{G M m}{x^2}\,dx = \dfrac{G M m}{r}$ , and $U(r) = U(\infty) - W = -\dfrac{G M m}{r}$ .
Equivalently, an outside agent must supply $+\dfrac{G M m}{r}$ to take $m$ back to infinity.
U belongs to the pair of masses, increases (becomes less negative) as r increases, and tends to 0 at infinity.
The force is conservative: the work depends only on the end points, not the path.

U(r) = -\dfrac{G M m}{r}

U = −G M m / r

U is negative everywhere and rises towards 0 as r increases.

Worked example

Worked example 1: a satellite’s potential energy

A $250$ kg satellite is at a distance $r = 1.0 \times 10^7$ m from the centre of the Earth ( $G = 6.67 \times 10^{-11}$ , $M_E = 6.0 \times 10^{24}$ kg). Find its gravitational potential energy and the gravitational potential there.

$U = -\dfrac{G M m}{r} = -\dfrac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 250}{1.0 \times 10^7} = -1.00 \times 10^{10}$ J.
$V = \dfrac{U}{m} = -\dfrac{G M}{r} = -4.00 \times 10^{7}$ J kg $^{-1}$ .

U = −G M m / r

U is negative everywhere and rises towards 0 as r increases.

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