Physics 9702 · Paper 4

Potential Energy, Orbit Changes & Escape

Multiply the potential by a mass and you get that mass's stored energy. Every orbit-energy calculation in Paper 4 starts here.

From potential to potential energy

E_p = m\phi = -\dfrac{GMm}{r}

Symbols

$E_p$ = gravitational potential energy (zero at infinity) (J)
$m$ = the mass placed in the field (kg)
$\phi$ = gravitational potential at the point (J kg⁻¹)

The link $E_p = m\phi$ is itself a one-marker (9702/42/O/N/23 Q2(b)(i)).
Like the potential, $E_p$ is negative everywhere: a satellite is trapped in the well until something gives it enough energy to reach zero.
The old $\Delta E_p = mg\Delta h$ from lesson 5.04 is the small-height version, valid only while g is unchanged. Between orbits, use the full formula.

Worked example

Exam version: energy to lift an orbit

A 122 kg satellite is moved from an orbit 1.7 × 10⁶ m above Mars's surface to one 6.8 × 10⁶ m above it. Mars has radius 3.4 × 10⁶ m, and the energy needed is 5.1 × 10⁴ kJ. Show that Mars's mass is about 6.4 × 10²³ kg. (9702/41/M/J/25 Q1(b)(i))

Radii measured from the centre: $r_1 = 3.4 + 1.7 = 5.1 \times 10^{6}$ m and $r_2 = 3.4 + 6.8 = 10.2 \times 10^{6}$ m. (The mark scheme awards a mark for this addition alone.)
$\Delta E_p = GMm\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right) = 5.1 \times 10^{8} \text{ J}$ .
Solve for M: $M = \dfrac{5.1 \times 10^{8}}{6.67 \times 10^{-11} \times 122 \times (1/5.1 - 1/10.2) \times 10^{-6}} = 6.4 \times 10^{23} \text{ kg}$ .

Answer

M = 6.4 \times 10^{23} \text{ kg}

Common mistake

Heights above a surface are not r. Add the planet's radius before using any formula in this chapter, and subtract it back if the question asks for a height. Both directions of this step carry their own mark (9702/42/M/J/23 Q1(c)).

Leaving the well: escape and one-way trips

To infinity from an orbit: the energy needed is the size of the potential energy minus the kinetic energy the satellite already has: $E = \dfrac{GMm}{r} - \tfrac{1}{2}mv^2$ (9702/42/F/M/24 Q1(b)(iii)).
Escape speed from a surface: launch fast enough that the kinetic energy pays the whole climb: $\tfrac{1}{2}mv^2 + m\phi = 0$ , so $v = \sqrt{-2\phi}$ (9702/42/O/N/23 Q2(b)(ii)).

Worked example

One change: escape speed from the Moon

The potential at the Moon's surface is −2.9 × 10⁶ J kg⁻¹. Find the escape speed. (9702/42/O/N/23 Q2(c))

$v = \sqrt{-2\phi} = \sqrt{2 \times 2.9 \times 10^{6}} = 2400 \text{ m s}^{-1}$ .

Answer

v = 2400 \text{ m s}^{-1}

Worked example

The reverse trip: falling from far away

A 200 kg rock, at rest very far from a planet (M = 4.9 × 10²³ kg, radius 3.4 × 10⁶ m), falls and hits the surface. Find its impact speed (9702-style, from the coursebook).

GPE lost = size of $\dfrac{GMm}{r} = \dfrac{6.67 \times 10^{-11} \times 4.9 \times 10^{23} \times 200}{3.4 \times 10^{6}} = 1.9 \times 10^{9} \text{ J}$ .
All of it becomes kinetic energy: $v = \sqrt{\dfrac{2 \times 1.92 \times 10^{9}}{200}} = 4400 \text{ m s}^{-1}$ .

Answer

v = 4400 \text{ m s}^{-1}

PreviousGravitational Potential NextGraphs of g and φ

Potential Energy, Orbit Changes & Escape

Multiply the potential by a mass and you get that mass's stored energy. Every orbit-energy calculation in Paper 4 starts here.

From potential to potential energy

E_p = m\phi = -\dfrac{GMm}{r}

Symbols

$E_p$ = gravitational potential energy (zero at infinity) (J)
$m$ = the mass placed in the field (kg)
$\phi$ = gravitational potential at the point (J kg⁻¹)

The link $E_p = m\phi$ is itself a one-marker (9702/42/O/N/23 Q2(b)(i)).
Like the potential, $E_p$ is negative everywhere: a satellite is trapped in the well until something gives it enough energy to reach zero.
The old $\Delta E_p = mg\Delta h$ from lesson 5.04 is the small-height version, valid only while g is unchanged. Between orbits, use the full formula.

Worked example

Exam version: energy to lift an orbit

Radii measured from the centre: $r_1 = 3.4 + 1.7 = 5.1 \times 10^{6}$ m and $r_2 = 3.4 + 6.8 = 10.2 \times 10^{6}$ m. (The mark scheme awards a mark for this addition alone.)
$\Delta E_p = GMm\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right) = 5.1 \times 10^{8} \text{ J}$ .
Solve for M: $M = \dfrac{5.1 \times 10^{8}}{6.67 \times 10^{-11} \times 122 \times (1/5.1 - 1/10.2) \times 10^{-6}} = 6.4 \times 10^{23} \text{ kg}$ .

Answer

M = 6.4 \times 10^{23} \text{ kg}

Common mistake

Leaving the well: escape and one-way trips

To infinity from an orbit: the energy needed is the size of the potential energy minus the kinetic energy the satellite already has: $E = \dfrac{GMm}{r} - \tfrac{1}{2}mv^2$ (9702/42/F/M/24 Q1(b)(iii)).
Escape speed from a surface: launch fast enough that the kinetic energy pays the whole climb: $\tfrac{1}{2}mv^2 + m\phi = 0$ , so $v = \sqrt{-2\phi}$ (9702/42/O/N/23 Q2(b)(ii)).

Worked example

One change: escape speed from the Moon

The potential at the Moon's surface is −2.9 × 10⁶ J kg⁻¹. Find the escape speed. (9702/42/O/N/23 Q2(c))

$v = \sqrt{-2\phi} = \sqrt{2 \times 2.9 \times 10^{6}} = 2400 \text{ m s}^{-1}$ .

Answer

v = 2400 \text{ m s}^{-1}

Worked example

The reverse trip: falling from far away

A 200 kg rock, at rest very far from a planet (M = 4.9 × 10²³ kg, radius 3.4 × 10⁶ m), falls and hits the surface. Find its impact speed (9702-style, from the coursebook).

GPE lost = size of $\dfrac{GMm}{r} = \dfrac{6.67 \times 10^{-11} \times 4.9 \times 10^{23} \times 200}{3.4 \times 10^{6}} = 1.9 \times 10^{9} \text{ J}$ .
All of it becomes kinetic energy: $v = \sqrt{\dfrac{2 \times 1.92 \times 10^{9}}{200}} = 4400 \text{ m s}^{-1}$ .

Answer

v = 4400 \text{ m s}^{-1}