Changing orbits

Builds on: Sections 3.2 and 3.3
Reading time: 40 min
Exercises: 4
Olympiad problems: 2

Spacecraft for Mars leave Earth in short launch windows that open about once every 26 months, and the cheapest path takes them about 8.5 months to get there. Both numbers come from orbits. The path is half of an ellipse that touches Earth's orbit at one end and Mars's at the other, and Mars must be in the right place to meet the spacecraft when it arrives. In this section we see how a short burn of a rocket engine changes an orbit, and how two such burns carry a spacecraft from one planet to another.

3.5.1A burn at one point

Before you readMake a guess first

A satellite on a circular orbit fires its engine forwards for a few seconds. Where on its new orbit is the point where it fired?

A rocket engine usually fires for a few minutes at most, much less than the period of an orbit, so we treat each burn as an instant change of velocity $Δ v$ at one point. The body stays at the same place, with a new velocity, and its new orbit must pass through that place. The vis-viva equation ((3.5)) then gives the new semi-major axis from the new speed.

A burn along the direction of motion at the pericentre raises the apocentre, and a burn against it lowers the apocentre. A burn at the apocentre changes the pericentre in the same way. The point of the burn always stays on the new orbit, so an orbit cannot be moved away from a point by firing there. To change a circular orbit into a bigger circle needs two burns, one at each end of the ellipse between them.

Example 3.8Raising the far side of an orbit

A satellite is on a circular orbit 300 km above the ground. It fires its engine forwards and gains 100 m/s. How high is its new apogee? Take $G M = 3.986 \times 10^{14}$ m³ s⁻² and Earth's radius as 6371 km.

Solution

The orbit's radius is 6671 km, and the speed of a circular orbit there is $\sqrt{G M / r} = 7.730$ km/s, so after the burn the satellite moves at 7.830 km/s. By (3.5),

\frac{1}{a} = \frac{2}{r} - \frac{v^{2}}{G M},

which gives $a = 6849$ km. The burn point is the perigee, so the apogee is $2 a - r = 7028$ km from Earth's centre.

AnswerThe new apogee is 657 km above the ground, on the far side of Earth from the burn. The perigee stays at 300 km.

3.5.2Small kicks

When the change of speed is small, the change of the orbit follows from the change of energy. On a circular orbit, a small forward kick $δ v$ raises the energy by $m v δ v$ , and since $E = - G M m / 2 a$ the semi-major axis grows in proportion. A small kick straight outwards or inwards changes the energy only very slightly, so it leaves the size of the orbit almost unchanged and makes it a little elliptical.

ResultSmall kicks on a circular orbit(3.12)

\begin{aligned} \frac{δ a}{a} & = 2 \frac{δ v}{v}, \frac{δ T}{T} = 3 \frac{δ v}{v} \\ e & = \frac{δ v_{r}}{v} \end{aligned}

$δ v$	a small change of speed along the direction of motion
$δ v_{r}$	a small change of speed along the radius, outwards or inwards
$v, a, T$	the speed, the radius and the period of the circular orbit before the kick
$e$	the eccentricity of the orbit after a kick along the radius

The first line comes from differentiating the vis-viva equation, and the period follows from

T \propto a^{3 / 2}

. A forward kick raises the far side of the orbit by

2 δ a = 4 a δ v / v

and leaves the kick point where it is.

For the burn of the example, $δ v / v = 0.1 / 7.730$ , and the far side rises by $4 \times 6671 \times 0.1 / 7.730 = 345$ km, to 645 km above the ground. The exact answer is 657 km, so the small-kick rule is already close for a change of speed of 1.3 per cent. For the International Space Station, moving at 7.66 km/s, every 1 m/s added in the direction of motion raises the far side of its orbit by about 3.5 km.

3.5.3The Hohmann transfer

The cheapest way from one circular orbit to another in the same plane, using two burns, is the half-ellipse that touches the inner orbit at one end and the outer orbit at the other. It is called a Hohmann transfer, after Walter Hohmann, who described it in 1925. Its semi-major axis is the average of the two radii, and the flight takes half its period.

ResultThe Hohmann transfer(3.13)

\begin{aligned} a_{t} & = \frac{r_{1} + r_{2}}{2}, t = \frac{T_{t}}{2} = π \sqrt{\frac{a_{t}^{3}}{G M}} \\ Δ v_{1} & = \sqrt{\frac{G M}{r_{1}}} (\sqrt{\frac{2 r_{2}}{r_{1} + r_{2}}} - 1) \\ Δ v_{2} & = \sqrt{\frac{G M}{r_{2}}} (1 - \sqrt{\frac{2 r_{1}}{r_{1} + r_{2}}}) \end{aligned}

$r_{1}, r_{2}$	the radii of the inner and the outer circular orbits
$a_{t}, T_{t}$	the semi-major axis and the period of the transfer ellipse
$t$	the time of the transfer
$Δ v_{1}, Δ v_{2}$	the changes of speed at the inner and the outer orbit
$M$	the mass of the central body

Each change of speed is the difference between the speed on the transfer ellipse, from (3.5), and the speed on the circle at the same point. Going outwards both burns are forwards. Going inwards both are backwards, with the same sizes.

Figure 3.4A Hohmann transfer to another planet

Target

Transfer flown

Watch

Transfer time
Changes of speed
At launch

The figure shows the Sun S and the orbits of Earth and of the target planet, drawn dotted and taken as circles in one plane, to scale. The heavy half-ellipse is the transfer, from the launch point L on Earth's orbit to the meeting point R on the planet's orbit, and its other half is dashed. As the spacecraft X flies, Earth E and the planet move on their circles at their true speeds, and the planet reaches R just as the spacecraft does. The table gives the time and the changes of speed from (3.13), and how far ahead of Earth the planet must be at launch. The figure can be set to , , or , an inner planet, which must start behind Earth.

Example 3.9From Earth to Mars

Take the orbits of Earth and Mars as circles of radius 1 au and 1.524 au. How long does a Hohmann transfer take, what changes of speed does it need, and where must Mars be at launch?

Solution

By (3.13), $a_{t} = (1 + 1.524) / 2 = 1.262$ au, so the transfer takes half of ${1.262}^{3 / 2}$ years, which is 259 days. In astronomical units with $\sqrt{G M / 1 au} = 29.78$ km/s,

\begin{aligned} Δ v_{1} & = 29.78 \times (\sqrt{1.2076} - 1) = 2.95 km/s \\ Δ v_{2} & = 24.12 \times (1 - \sqrt{0.7924}) = 2.65 km/s, \end{aligned}

where 24.12 km/s is the speed of Mars on its orbit. Mars takes 687.2 days to go round, so in the 258.9 days of the transfer it moves $360 ° \times 258.9 / 687.2 = 135.6 °$ . It must arrive at R, 180° round from L, so at launch it must be $180 ° - 135.6 ° = 44.4 °$ ahead of Earth.

AnswerThe transfer takes 259 days, about 8.5 months, with changes of speed of 2.95 km/s at Earth's orbit and 2.65 km/s at Mars's. At launch Mars must be 44.4° ahead of Earth.

Quick checkHow often the window opens

Mars goes round the Sun in 1.881 years. How often does Mars return to the same position relative to Earth, so that a launch window to Mars opens again?

days

Common errorsSection 3.5

A single burn cannot move an orbit away from the burn point. To go from one circle to a larger one takes two burns, the second at the far end of the transfer ellipse.
The change of speed at each end is the difference between the speeds on the transfer ellipse and on the circle at the same point. The speed of the circle at one end must not be compared with the speed of the ellipse at the other.
A forward burn makes a satellite move more slowly in the end, on its new, higher orbit. To catch up with something ahead on the same orbit, a spacecraft first slows down, drops to a lower and faster orbit, and then speeds up again.
When the apocentre of a long, thin orbit comes from a small difference, such as $2 G M / (v_{p}^{2} r_{p}) - 1$ , every rounding along the way is magnified. Four significant figures should be kept until the end.

ExercisesSection 3.54 questions

3.5.1
A satellite on a circular orbit speeds up by 1 per cent with a short forward burn. By about how many per cent does its period change?
per cent
3.5.2
A satellite moving at 7.66 km/s on a circular orbit gets a kick of 0.1 km/s straight outwards from Earth. What is the eccentricity of its new orbit?
3.5.3
A spacecraft is to go from Earth to Venus by a Hohmann transfer. What must it do at the start?
3.5.4
How long does a Hohmann transfer from Earth, at 1 au, to Jupiter, at 5.203 au, take?
years

Olympiad problemsPast IOAA problems, restated2 problems

2016Raising the orbit of the Mars Orbiter Mission20 points

Adapted from IOAA 2016 (Bhubaneswar), theory problem 9. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

India's Mars Orbiter Mission (MOM) was launched on 5 November 2013. Its dry mass, the body and the instruments, was 500 kg, and it carried 852 kg of fuel. It was first placed on an elliptical orbit round Earth with its perigee 264.1 km and its apogee 23 903.6 km above the surface. After six raisings of the orbit it was sent on its transfer orbit to Mars.

The first raising was made by firing the engines for a very short time near perigee, without changing the plane of the orbit or its perigee. This gave the spacecraft an impulse of $1.73 \times 10^{5}$ kg m s⁻¹. Ignore the loss of mass as the fuel burned. Take Earth's radius as 6371 km, its mass as $5.972 \times 10^{24}$ kg and $G = 6.674 \times 10^{- 11}$ N m² kg⁻².

(a) What is the height $h_{a}$ of the new apogee above Earth's surface? [14 points]

(b) Find the eccentricity of the new orbit and its period in hours. [6 points]

I solved it without the solution

Answer(a) The new apogee is about 28 860 km above the surface. (b) The eccentricity is 0.683 and the period 8.37 hours. The official solution gives 29 009 km, about 150 km higher, because it rounds the speeds and the ratio along the way. The apogee depends on the small difference 0.1883, so every rounding is magnified, and we keep four figures until the end.

2011Which orbits can one kick reach?10 points

Adapted from IOAA 2011 (Katowice), theory problem 7. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

A satellite moves on a circular orbit round Earth with momentum $p$ . An explosive charge gives it an extra impulse $Δ p$ equal in size to $p$ . Let $α$ be the angle between $p$ and $Δ p$ , and $β$ the angle between the satellite's position vector from Earth's centre and $Δ p$ . For each orbit below, say whether it can be reached, and if so give values of $α$ and $β$ that reach it. For example, $α = 180 °$ and $β = 90 °$ leave the satellite at rest, so that it falls straight down.

(a) A hyperbola with its perigee at the point of the explosion.

(b) A parabola with its perigee at the point of the explosion.

(d) A circle.

(e) An ellipse with its apogee at the point of the explosion.

I solved it without the solution

AnswerAll five are possible, with $β = 90 °$ and (a) $α$ less than 90°, (b) $α = 90 °$ , (c) $α$ between 90° and 120°, (d) $α = 120 °$ , (e) $α$ between 120° and 180°. For any $α$ other than 0° and 180° the kick has a part out of the plane of the old orbit, so the new orbit is tilted.