Motion in Orbits

Lessons

Senior 5 · Physics

Motion in Orbits

4 lessons

Lesson 2 of 4

Variation of g and Gravitational Energy

25 min

Are you lighter in Kigali?

The acceleration of free fall is often quoted as $9.81\,\mathrm{m\,s^{-2}}$ , but that is a typical value, not a constant of nature. It is smaller on a mountain than at sea level, smaller at the equator than at the poles, and it changes as you go down a mine shaft. Kigali lies about 2° south of the equator at around 1567 m above sea level, so a pendulum there measures a little less than 9.81, close to $9.78\,\mathrm{m\,s^{-2}}$ , as Experiment 1.2 found. In this lesson you work out how $g$ changes with height and depth, and then turn to the energy needed to move masses in a gravitational field, up to the speed needed to leave the Earth for good.

g above the surface

At height $h$ above the surface, the distance from the Earth's centre is $R + h$ :

g_h = \frac{GM}{(R + h)^2} = g_0\frac{R^2}{(R + h)^2} = \frac{g_0}{\left(1 + \frac{h}{R}\right)^2}

where $g_0 = \frac{GM}{R^2}$ is the surface value. For heights much smaller than $R$ , the binomial approximation $(1 + x)^{-2} \approx 1 - 2x$ gives

g_h \approx g_0\left(1 - \frac{2h}{R}\right) \qquad (h \ll R)

Climbing 1567 m reduces $g$ by $9.81 \times \frac{2 \times 1567}{6.37 \times 10^6} = 0.0048\,\mathrm{m\,s^{-2}}$ : small, but measurable with a good pendulum or a modern gravimeter.

g below the surface

Go down a mine to a depth $d$ , so you are at $r = R - d$ from the centre. Newton showed that a uniform spherical shell exerts no net force on a mass inside it: the pulls from all parts of the shell cancel. So only the sphere of radius $r$ beneath you attracts you. If the Earth had uniform density $\rho$ , that inner sphere would have mass $M_r = \frac{4}{3}\pi r^3\rho = M\frac{r^3}{R^3}$ , and

g_d = \frac{GM_r}{r^2} = \frac{GM}{R^3}\,r = g_0\frac{r}{R} = g_0\left(1 - \frac{d}{R}\right)

Inside a uniform Earth, $g$ falls in a straight line to zero at the centre.

(The real Earth is much denser near its core, so in fact $g$ first rises slightly with depth before it falls; the uniform model is the one used in examinations.)

Rotation and shape

Two more effects make $g$ depend on latitude. The Earth spins, and part of the gravitational pull on a body at the equator provides its centripetal acceleration, $\omega^2R = (7.29 \times 10^{-5})^2 \times 6.37 \times 10^6 = 0.034\,\mathrm{m\,s^{-2}}$ ; at the poles this effect is zero. And the Earth bulges at the equator, so the equator is further from the centre. Together these make $g$ about $9.78\,\mathrm{m\,s^{-2}}$ at sea level on the equator and $9.83\,\mathrm{m\,s^{-2}}$ at the poles.

Worked example 1: g in orbit and underground

Taking $g_0 = 9.81\,\mathrm{m\,s^{-2}}$ and $R = 6.37 \times 10^6\,\mathrm{m}$ , find $g$ (a) at the height of the International Space Station, about 400 km, (b) at a depth of half the Earth's radius, assuming uniform density.

Solution

(a) $g = 9.81 \times \left(\frac{6.37}{6.77}\right)^2 = 9.81 \times 0.885 = 8.7\,\mathrm{m\,s^{-2}}$ . Gravity at the ISS is still 89% of its surface value: astronauts float because they are in free fall with their station, not because gravity has vanished.

(b) $g = g_0\frac{r}{R} = 9.81 \times 0.5 = 4.9\,\mathrm{m\,s^{-2}}$ .

Gravitational potential and potential energy

The familiar $\Delta E_p = mgh$ assumes $g$ is constant, which is fine for a flight of stairs but not for a rocket. To lift a mass $m$ from distance $r$ to infinity, you must do work against the force $\frac{GMm}{x^2}$ at every distance $x$ along the way:

W = \int_r^\infty \frac{GMm}{x^2}\,dx = \left[-\frac{GMm}{x}\right]_r^\infty = \frac{GMm}{r}

We choose the potential energy to be zero at infinity, where the bodies no longer interact. Since work $\frac{GMm}{r}$ must be done to get there, the potential energy at $r$ is below zero by that amount:

U = -\frac{GMm}{r}, \qquad V = \frac{U}{m} = -\frac{GM}{r}

$V$ is the gravitational potential, the potential energy per unit mass, in $\mathrm{J\,kg^{-1}}$ . It is negative everywhere and rises towards zero with distance: the Earth sits at the bottom of a "potential well".

The change in potential energy between two distances is

\Delta U = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)

Near the surface, with $r_{1} = R$ and $r_{2} = R + h$ , this is $\frac{GMmh}{R(R + h)} \approx \frac{GMm}{R^2}h = mgh$ . So $m g h$ is the small-height approximation of the general result.

Definition

The gravitational potential $V$ at a point is the work done per unit mass in bringing a small mass from infinity to that point: $V = -\frac{GM}{r}$ . The gravitational potential energy of a mass $m$ there is $U = mV = -\frac{GMm}{r}$ . Example: at the Earth's surface $V = -\frac{3.99 \times 10^{14}}{6.37 \times 10^6} = -6.26 \times 10^7\,\mathrm{J\,kg^{-1}}$ .

Worked example 2: is mgh good enough?

Find the increase in potential energy when a 1000 kg satellite is raised from the Earth's surface to a height of 400 km, (a) using $\Delta U = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$ , (b) using $m g h$ .

Solution

(a) $\Delta U = 3.99 \times 10^{14} \times 1000 \times \left(\frac{1}{6.37 \times 10^6} - \frac{1}{6.77 \times 10^6}\right) = 3.99 \times 10^{17} \times 9.28 \times 10^{-9} = 3.70 \times 10^9\,\mathrm{J}$ .

(b) $mgh = 1000 \times 9.81 \times 4.0 \times 10^5 = 3.92 \times 10^9\,\mathrm{J}$ , about 6% too large, because $g$ is smaller higher up. For heights that are a noticeable fraction of $R$ , use the full formula.

Escape velocity

A ball thrown upwards slows down and falls back. Throw it faster and it rises higher. Is there a speed at which it never comes back? It must reach infinity (where $U = 0$ ) with at least zero kinetic energy, so its total energy at launch must be at least zero:

\tfrac{1}{2}mv_e^2 - \frac{GMm}{R} = 0 \quad\Longrightarrow\quad v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2g_0R}

For the Earth, $v_e = \sqrt{2 \times 9.81 \times 6.37 \times 10^6} = 1.12 \times 10^4\,\mathrm{m\,s^{-1}} = 11.2\,\mathrm{km\,s^{-1}}$ . The escape velocity does not depend on the mass being launched, and ignores air resistance. It also explains atmospheres: gas molecules move at hundreds of metres per second, and on a small body with a low escape velocity, enough of them reach escape speed for the gas to leak away over time.

Worked example 3: escaping from the Moon

Find the escape velocity from the Moon ( $M = 7.35 \times 10^{22}\,\mathrm{kg}$ , $R = 1.74 \times 10^6\,\mathrm{m}$ ) and compare it with the Earth's.

Solution

$v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{1.74 \times 10^6}} = \sqrt{5.64 \times 10^6} = 2.37 \times 10^3\,\mathrm{m\,s^{-1}}$ , about a fifth of the Earth's. This low value helps explain why the Moon has kept almost no atmosphere, and why the Apollo astronauts needed only a small rocket to leave it.

Warning

Common mistakes. (1) Using $m g h$ for large heights, such as a satellite orbit. (2) Dropping the minus sign in $U = -\frac{GMm}{r}$ : potential energy increases (becomes less negative) as $r$ increases. (3) Using $g_0(1 - \frac{2h}{R})$ when $h$ is not small compared with $R$ ; then use the exact $\frac{R^2}{(R + h)^2}$ . (4) Thinking astronauts in orbit are weightless because there is no gravity there.

Tip

Above the surface $g \propto \frac{1}{r^2}$ ; below it (uniform Earth) $g \propto r$ . Sketch the two-piece graph before you calculate: it tells you whether the answer should be bigger or smaller than $g_{0}$ , and by roughly how much.

Gravitational potential energy and escape velocity (UNSW Physics). Watch why the potential energy is taken as zero at infinity and negative everywhere else, and how that leads to the escape speed.

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Practice

At what height above the Earth's surface is $g$ one quarter of its surface value?
Use the approximation $g_h \approx g_0(1 - \frac{2h}{R})$ to find $g$ on top of a mountain 4500 m high. Check with the exact formula.
Find $g$ at the bottom of a mine 3.0 km deep, assuming a uniform Earth. Is the change bigger or smaller than for a height of 3.0 km?
Show that $g_d = g_0(1 - \frac{d}{R})$ for a uniform Earth, stating the result about spherical shells that you use.
Find the gravitational potential at a height of $2 R$ above the Earth's surface.
How much energy is needed to raise a 500 kg satellite from the surface to a height of $1.0 \times 10^6\,\mathrm{m}$ ?
Derive the escape velocity from a planet of mass $M$ and radius $R$ . Calculate it for Mars ( $6.42 \times 10^{23}\,\mathrm{kg}$ , $3.39 \times 10^6\,\mathrm{m}$ ).
Mugisha says: "Escape velocity means a rocket must reach 11.2 km/s, or it can never leave the Earth." Evaluate this statement.

Show answersHide answers

$\frac{R^2}{(R + h)^2} = \frac{1}{4}$ gives $R + h = 2 R$ : $h = R = 6.37 \times 10^6\,\mathrm{m}$ .
$g \approx 9.81\left(1 - \frac{9000}{6.37 \times 10^6}\right) = 9.81 \times 0.99859 = 9.796\,\mathrm{m\,s^{-2}}$ . Exact: $9.81 \times \left(\frac{6.37}{6.3745}\right)^2 = 9.796\,\mathrm{m\,s^{-2}}$ : the same to four figures.
$g = 9.81 \times \left(1 - \frac{3.0}{6370}\right) = 9.805\,\mathrm{m\,s^{-2}}$ : a fall of 0.0046, half the fall for a height of 3.0 km (0.0092), because depth gives $\frac{d}{R}$ and height $\frac{2h}{R}$ .
A uniform shell exerts no net force on a mass inside it, so only the sphere of radius $r = R - d$ counts: $g = \frac{G(Mr^3/R^3)}{r^2} = g_0\frac{r}{R} = g_0\left(1 - \frac{d}{R}\right)$ .
$r = 3 R$ : $V = -\frac{GM}{3R} = -\frac{6.26 \times 10^7}{3} = -2.09 \times 10^7\,\mathrm{J\,kg^{-1}}$ .
$\Delta U = 3.99 \times 10^{14} \times 500 \times \left(\frac{1}{6.37 \times 10^6} - \frac{1}{7.37 \times 10^6}\right) = 4.25 \times 10^9\,\mathrm{J}$ .
$\frac{1}{2}mv^2 = \frac{GMm}{R}$ , so $v_e = \sqrt{\frac{2GM}{R}}$ . Mars: $\sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{3.39 \times 10^6}} = 5.0\,\mathrm{km\,s^{-1}}$ .
Not correct. Escape velocity is the launch speed needed by an unpowered projectile that gets no more push after leaving the ground. A rocket keeps firing its engines as it climbs, so it can leave the Earth while moving more slowly; what it needs in total is enough energy, not that particular speed at the surface.

Key points

Above the surface: $g_h = g_0\frac{R^2}{(R + h)^2} \approx g_0(1 - \frac{2h}{R})$ . Below (uniform Earth): $g_d = g_0(1 - \frac{d}{R})$ .
Rotation and the equatorial bulge make $g$ smaller at the equator than at the poles.
$V = -\frac{GM}{r}$ , $U = -\frac{GMm}{r}$ , zero at infinity; $m g h$ is its small-height approximation.
Escape velocity $v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} = 11.2\,\mathrm{km\,s^{-1}}$ for the Earth.

Sources and credits

13.2 Gravitation Near Earth's Surface and 13.3 Gravitational Potential Energy and Total Energy, from University Physics Volume 1 by William Moebs, Samuel J. Ling and Jeff Sanny, OpenStax, licensed under CC BY 4.0. Adapted by Ganzaa.
Kigali's latitude and elevation: Kigali, Wikipedia, licensed under CC BY-SA 4.0; facts only. The value of about $9.78\,\mathrm{m\,s^{-2}}$ is calculated by Ganzaa from the international gravity formula with a height correction.
Video: Gravitational potential energy & Escape Velocity, UNSW Physics, embedded from YouTube.

The two diagrams, the worked examples and the practice exercises are written by Ganzaa.

1Are you lighter in Kigali?

2g above the surface

3g below the surface

Rotation and shape

4Gravitational potential and potential energy

5Escape velocity

Are you lighter in Kigali?

g above the surface

g below the surface

Gravitational potential and potential energy

Escape velocity