Cosmic Velocities and Satellites – Motion in Orbits (Senior 5 Physics) | Ganzaa

Motion in Orbits

Lessons

Senior 5 · Physics

Motion in Orbits

4 lessons

Lesson 4 of 4

Cosmic Velocities and Satellites

25 min

Newton's cannon

Newton imagined a cannon on top of a very high mountain, firing horizontally. A slow shot curves down and lands nearby; a faster one lands further away. But the Earth is round, so the ground curves away beneath the shot as it falls. Fire it fast enough and the ground curves away exactly as fast as the shot falls: it never lands, and keeps circling the Earth. That shot is a satellite. In 2019, Rwanda's first satellite, RwaSat-1, began exactly this endless fall; it was launched through the Japan Aerospace Exploration Agency (JAXA) and gathered data on soil moisture for estimating crop yields. This lesson finds the speeds that put a body into orbit, free it from the Earth, and free it from the Sun.

The first cosmic velocity

For a circular orbit just above the surface (ignoring air resistance and mountains), gravity provides the centripetal force at radius $R$ :

\frac{mv_1^2}{R} = \frac{GMm}{R^2} \quad\Longrightarrow\quad v_1 = \sqrt{\frac{GM}{R}} = \sqrt{gR} = \sqrt{9.81 \times 6.37 \times 10^6} = 7.9\,\mathrm{km\,s^{-1}}

This is the first cosmic velocity, the least speed for a circular orbit around the Earth. Such an orbit would take $T = \frac{2\pi R}{v_1} = 84$ minutes. Real satellites orbit above the atmosphere, at larger $r$ , and so a little more slowly: $v = \sqrt{\frac{GM}{r}}$ .

The second cosmic velocity

Between $v_{1}$ and a second value, the shot follows an ellipse that swings further out on the far side of the Earth and comes back. From Lesson 2, a body escapes completely if its total energy is at least zero:

v_2 = \sqrt{\frac{2GM}{R}} = \sqrt{2}\,v_1 = 11.2\,\mathrm{km\,s^{-1}}

This second cosmic velocity is the escape velocity. At exactly $v_{2}$ the path is a parabola; faster still, a hyperbola.

The third cosmic velocity

A body that has escaped from the Earth is still held by the Sun. At the Earth's distance ( $r = 1.496 \times 10^{11}\,\mathrm{m}$ ), the speed needed to escape from the Sun is $\sqrt{\frac{2GM_S}{r}} = 42.1\,\mathrm{km\,s^{-1}}$ . The Earth already carries everything on it round the Sun at $\sqrt{\frac{GM_S}{r}} = 29.8\,\mathrm{km\,s^{-1}}$ , so a probe launched in the direction of the Earth's motion needs only an extra $42.1 - 29.8 = 12.3\,\mathrm{km\,s^{-1}}$ once it is clear of the Earth. Launching from the surface, it must also climb out of the Earth's field. By energy conservation, per unit mass,

\tfrac{1}{2}v_3^2 = \tfrac{1}{2}v_2^2 + \tfrac{1}{2}(12.3)^2 \quad\Longrightarrow\quad v_3 = \sqrt{11.2^2 + 12.3^2} = 16.6\,\mathrm{km\,s^{-1}}

The usual quoted value of this third cosmic velocity is $16.7\,\mathrm{km\,s^{-1}}$ , the least launch speed from the Earth's surface to leave the Solar System.

Definition

The cosmic velocities for the Earth are: the first, $v_1 = \sqrt{gR} \approx 7.9\,\mathrm{km\,s^{-1}}$ , the speed for a circular orbit at the surface; the second (escape velocity), $v_2 = \sqrt{2gR} \approx 11.2\,\mathrm{km\,s^{-1}}$ , the least speed to escape the Earth's gravity; the third, $v_3 \approx 16.7\,\mathrm{km\,s^{-1}}$ , the least launch speed from the Earth to escape the Sun's gravity. All ignore air resistance.

Worked example 1: the International Space Station

The ISS orbits about 400 km above the surface. Find its speed and period, and the number of orbits it makes each day. ( $GM = 3.99 \times 10^{14}\,\mathrm{m^3\,s^{-2}}$ , $R = 6.37 \times 10^6\,\mathrm{m}$ .)

Solution

$r = 6.77 \times 10^6\,\mathrm{m}$ ; $v = \sqrt{\frac{3.99 \times 10^{14}}{6.77 \times 10^6}} = 7.67\,\mathrm{km\,s^{-1}}$ . $T = \frac{2\pi r}{v} = \frac{2\pi \times 6.77 \times 10^6}{7.67 \times 10^3} = 5.55 \times 10^3\,\mathrm{s} = 92\,\mathrm{min}$ . Orbits per day: $\frac{1440}{92} \approx 15.6$ . The astronauts see about 16 sunrises a day.

Satellites and their orbits

A satellite is any body that orbits a larger one: the Moon is a natural satellite of the Earth, and thousands of artificial satellites now circle it. Three kinds of orbit are especially useful.

Low Earth orbits (a few hundred to about 2000 km up): periods of 90 to 120 minutes. Close to the ground, they suit Earth observation, internet constellations and the ISS. A polar orbit passes over both poles; as the Earth turns beneath it, the satellite eventually passes over every place, which is ideal for mapping, weather and crop monitoring.
Navigation orbits: GPS satellites orbit at about 20 200 km with periods of about 12 hours; a receiver times signals from at least four of them to find its position.
Geostationary orbit: a circular orbit above the equator, moving in the same direction as the Earth spins, with a period equal to one rotation of the Earth (23 h 56 min). The satellite then stays above the same point on the equator, so a fixed dish can point at it.

The radius of a geostationary orbit

From $T^2 = \frac{4\pi^2}{GM}r^3$ with $T = 86\,164\,\mathrm{s}$ :

r = \left(\frac{GMT^2}{4\pi^2}\right)^{\frac{1}{3}} = \left(\frac{3.99 \times 10^{14} \times 86\,164^2}{4\pi^2}\right)^{\frac{1}{3}} = 4.22 \times 10^7\,\mathrm{m}

That is about 42 200 km from the Earth's centre, or 35 800 km above the equator, with a speed of $\frac{2\pi r}{T} = 3.07\,\mathrm{km\,s^{-1}}$ . Every geostationary satellite in the world shares this one ring.

Use	Typical orbit	Example
television broadcasting and some communication	geostationary	satellite TV dishes on homes point at a fixed spot in the sky
navigation	about 20 200 km, 12 h	GPS and other navigation systems in phones
weather	geostationary and polar	cloud images used in daily forecasts
Earth observation and farming	low, often polar	RwaSat-1 measured soil moisture to help estimate crop yields
internet	low Earth orbit constellations	Rwanda's second satellite, Icyerekezo, helped bring internet to Nkombo Island on Lake Kivu

Worked example 2: comparing two orbits

Without using $G$ or $M$ , find the period of a satellite whose orbit radius is 4.0 times that of a satellite with period 90 min.

Solution

$\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{\frac{3}{2}} = 4.0^{1.5} = 8.0$ , so $T_2 = 8.0 \times 90 = 720\,\mathrm{min} = 12\,\mathrm{h}$ : roughly a GPS orbit.

Worked example 3: why astronauts float

An astronaut of mass 70 kg is in the ISS, where $g = 8.7\,\mathrm{N\,kg^{-1}}$ . Find the gravitational force on her, and explain why a bathroom scale under her feet would read zero.

Solution

$F = mg = 70 \times 8.7 = 610\,\mathrm{N}$ : gravity is still strong. But she and the station both fall towards the Earth with the same acceleration, 8.7 m s⁻², so the floor does not push on her and the scale reads zero. "Weightlessness" in orbit is continuous free fall, the same feeling as the first moment of a jump, not an absence of gravity.

Warning

Common mistakes. (1) Saying satellites stay up because there is no gravity in space. (2) Using the height above the surface as $r$ in $v = \sqrt{\frac{GM}{r}}$ : $r$ is from the Earth's centre. (3) Thinking any 24-hour orbit is geostationary: it must also be circular, above the equator, and in the direction of the Earth's spin. (4) Mixing up the cosmic velocities: 7.9 (orbit), 11.2 (escape the Earth), 16.7 km/s (escape the Sun from the Earth).

Tip

Remember the ratio $v_2 = \sqrt{2}\,v_1$ : $7.9 \times 1.41 = 11.2\,\mathrm{km\,s^{-1}}$ . And for any orbit, kinetic energy $= \frac{GMm}{2r}$ is exactly half the energy needed to escape from that orbit.

Newton's Cannon walkthrough (CK-12 Foundation). Watch how raising the launch speed turns a falling shot into a circular orbit, then an ellipse, then an escape.

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Practice

Derive the first cosmic velocity and calculate it for the Earth.
Show that the second cosmic velocity is $\sqrt{2}$ times the first.
Calculate the first and second cosmic velocities for Mars ( $g = 3.73\,\mathrm{N\,kg^{-1}}$ , $R = 3.39 \times 10^6\,\mathrm{m}$ ).
A satellite orbits 800 km above the Earth. Find its speed and period.
Calculate the height above the surface of a geostationary satellite, and explain why all such satellites lie above the equator.
A weather satellite in a polar orbit has a period of 100 min. How many times does it pass over the equator each day, and why is this orbit useful for imaging all of Rwanda?
Explain why launching rockets eastwards, near the equator, saves fuel.
Keza says: "To reach a geostationary orbit, a rocket must reach 11.2 km/s, the escape velocity." Evaluate this statement.

Show answersHide answers

$\frac{mv^2}{R} = \frac{GMm}{R^2} = mg$ , so $v_1 = \sqrt{gR} = \sqrt{9.81 \times 6.37 \times 10^6} = 7.9\,\mathrm{km\,s^{-1}}$ .
$v_2 = \sqrt{\frac{2GM}{R}} = \sqrt{2}\sqrt{\frac{GM}{R}} = \sqrt{2}\,v_1$ .
$v_1 = \sqrt{3.73 \times 3.39 \times 10^6} = 3.56\,\mathrm{km\,s^{-1}}$ ; $v_2 = 5.03\,\mathrm{km\,s^{-1}}$ .
$r = 7.17 \times 10^6\,\mathrm{m}$ ; $v = \sqrt{\frac{3.99 \times 10^{14}}{7.17 \times 10^6}} = 7.46\,\mathrm{km\,s^{-1}}$ ; $T = \frac{2\pi r}{v} = 6.04 \times 10^3\,\mathrm{s} = 101\,\mathrm{min}$ .
$h = 4.22 \times 10^7 - 6.37 \times 10^6 = 3.58 \times 10^7\,\mathrm{m}$ (35 800 km). The centre of every orbit must be the Earth's centre; only an orbit in the equatorial plane can stay over one point as the Earth spins.
$\frac{1440}{100} = 14.4$ orbits a day, each crossing the equator twice: about 29 crossings. As the Earth turns beneath the orbit, successive passes cover different strips, so every part of the country is imaged regularly.
The Earth's surface at the equator already moves eastwards at about $\frac{2\pi \times 6.37 \times 10^6}{86\,164} = 465\,\mathrm{m\,s^{-1}}$ ; a rocket launched east keeps this speed, so it needs less extra speed to reach orbit.
Wrong. A geostationary satellite is still bound to the Earth (its total energy is negative), so it needs less energy than escape. In energy terms, it needs $GMm\left(\frac{1}{R} - \frac{1}{2r}\right)$ , which is less than the escape energy $\frac{GMm}{R}$ , and rockets gain this energy gradually rather than as one surface speed.

Key points

First cosmic velocity $v_1 = \sqrt{gR} = 7.9\,\mathrm{km\,s^{-1}}$ ; second $v_2 = \sqrt{2gR} = 11.2\,\mathrm{km\,s^{-1}}$ ; third $v_3 \approx 16.7\,\mathrm{km\,s^{-1}}$ .
Orbital speed $v = \sqrt{\frac{GM}{r}}$ ; higher orbits are slower and longer.
Geostationary: equatorial, eastward, $T = 23\,\mathrm{h}\,56\,\mathrm{min}$ , $r = 42\,200\,\mathrm{km}$ .
Satellites serve communication, navigation, weather, Earth observation and internet; RwaSat-1 (2019) monitored soil moisture.

1Newton's cannon

2The first cosmic velocity

3The second cosmic velocity

4The third cosmic velocity

5Satellites and their orbits

The radius of a geostationary orbit

Newton's cannon

The first cosmic velocity

The second cosmic velocity

The third cosmic velocity

Satellites and their orbits