Orbital Maneuvers

Guide

Orbital Maneuvers

Orbital maneuvers are the use of propulsion systems to change the orbit of a spacecraft. Orbital maneuvers are used to transfer a spacecraft from one orbit to another, to change the orientation of the orbit, or to change the speed of the spacecraft. The most common type of orbital maneuver is the Hohmann transfer orbit, which is used to transfer a spacecraft between two circular orbits of different altitudes around a central body. Other types of orbital maneuvers include bi-elliptic transfers, gravity assists, and plane changes. A wonderful resource to learn more about orbital mechanics is this book and this thing.

Delta-v

Delta-v ( $\Delta v$ ) is a measure of the impulse per unit of spacecraft mass that is needed to perform a maneuver. It is a scalar that has the units of speed. In the absence of external forces, the Delta-v is equal to the change in velocity of the spacecraft.

\tag{3.6.1} \Delta v = \int_{t_0}^{t_1} |\dot{v}| dt

where $\dot{v}$ is the acceleration. When the thrust force is applied in a constant direction, Delta-v is given by $\Delta v = |v_1 - v_0|$ , which is simply the magnitude of the change in velocity. For multiple maneuvers, Delta-v sums linearly.

Orbit maneuvers are made by firing a thruster to produce a reaction force acting on the spacecraft. The size of this force is

\tag{3.6.2} T = v_\text{exh} \, \rho

where

$v_\text{exh}$ is the velocity of the exhaust gas in rocket frame
$\rho$ is the mass flow rate of the propellant

If the mass of the satellite at $t=t_0$ before the maneuver is $m_0$ , and at $t=t_1$ after the maneuver is $m_1$ , then the Delta-v is given by the Tsiolkovsky rocket equation

\tag{3.6.3} \boxed{\Delta v = v_\text{exh} \ln \left( \frac{m_0}{m_1} \right)}

Thus the amount of propellant required for the maneuver is $m_1 - m_0$ .

When applying Delta-v in the direction of the velocity the specific orbital energy gained per unit Delta-v is equal to the instantaneous speed. This is called the Oberth effect. For example, a satellite in an elliptical orbit is boosted more efficiently at high speed (at perigee) than at low speed (at apogee). Another example is that when a vehicle is making a pass of a planet, burning the propellant at closest approach rather than further out gives significantly higher final speed, and this is even more so when the planet is a large one with a deep gravity field, such as Jupiter.

Hohmann Transfer Orbital Maneuver

The Hohmann transfer orbit is an orbital maneuver used to transfer a spacecraft between two orbits of different altitudes around a central body. The initial and target orbits are both circular and coplanar. The maneuver is accomplished by placing the craft into an elliptical transfer orbit that is tangential to both the initial and target orbits. The maneuver uses two impulsive engine burns: the first establishes the transfer orbit, and the second adjusts the orbit to match the target. The Hohmann maneuver often uses the lowest possible amount of impulse (which consumes a proportional amount of delta-v, and hence propellant).

The Hohmann transfer orbit is an orbital maneuver used to transfer a spacecraft between two orbits of different altitudes around a central body. (Source: Wikipedia)

Transfer between circular coplanar orbits

Consider an orbital transfer from orbit 1, having radius $a_1$ to orbit 2, having radius $a_2$ , via an elliptical orbit whose periapsis is in orbit 1 and apoapsis in orbit 2. Such transfer orbits are known as Hohmann least energy two-impulse cotangential orbits. The first impulse is applied at periapsis, and the second at apoapsis.

The semi-major axis of the transfer orbit is

\tag{3.6.4} a_t = \frac{a_1 + a_2}{2}

The eccentricity of the transfer orbit is

\tag{3.6.5} e_t = \frac{a_2 - a_1}{a_2 + a_1}

The transfer time is half the period of the transfer orbit, which is given by

\tag{3.6.6} T_t = \pi \sqrt{\frac{a_t^3}{\mu}}

If we work in the units of AU, years and solar masses, $T_t = \sqrt{\frac{(a_1 + a_2)^3}{32M}}$ .

The velocity at periapsis of the transfer orbit is greater than the velocity of the initial orbit, and the velocity at apoapsis of the transfer orbit is less than the velocity of the final orbit. The change in velocity at periapsis is given by

\tag{3.6.7} \Delta v_1 = v_\text{p,t} - v_1 = \sqrt{\frac{\mu}{a_t} \left( \frac{1 + e_t}{1 - e_t} \right) } - \sqrt{\frac{\mu}{a_1}}

The change in velocity at apoapsis is given by

\tag{3.6.8} \Delta v_2 = v_2 - v_\text{a,t} = \sqrt{\frac{\mu}{a_2}} - \sqrt{\frac{\mu}{a_t} \left( \frac{1 - e_t}{1 + e_t} \right) }

The total Delta-v for the maneuver is

\tag{3.6.9} \Delta v = \sqrt{\frac{\mu}{a_t (1 - e_t^2)}} \, (2e_t + \sqrt{1 - e_t} - \sqrt{1 + e_t})

Coplanar rendezvous transfer

The Hohmann transfer orbit can be used to rendezvous with another spacecraft in a co-planar orbit. Let’s say we have an interceptor satellite which has to be maneuvered to rendezvous with a target satellite. The interceptor is in a circular orbit of radius $a_1$ , and the target is in a circular orbit of radius $a_2$ . Both orbit in the same direction. The interceptor will perform a Hohmann transfer to reach the target orbit, and then it will perform a second maneuver to match the velocity of the target. The properties of the orbits are the same as in the previous section. The only difference is that the target is moving in its orbit, and the interceptor has to catch up to it, at just the right time.

A launch window is a time period during which the interceptor can be launched to reach the target. The launch window is determined by the relative positions of the two satellites in their orbits. The launch window is open when the interceptor is at periapsis, and the target is at apoapsis.

Let’s assume that the interceptor and target are in just the right configuration for the rendezvous maneuver. The lead angle $\alpha_\text{lead}$ is the angle between the initial position of the target and the (future) rendezvous point.

\tag{3.6.10} \alpha_\text{lead} = \omega_\text{target} \cdot T_t

where $\omega_\text{target}$ is the angular velocity of the target, and $T_t$ is the time it takes for the interceptor to reach the target orbit. The phase angle $\phi$ is the angle between the interceptor and the target at any instant. It is measured from the interceptor to the target, in the direction of motion. At the start of the maneuver, the phase angle is given by

\tag{3.6.11} \phi_0 = \pi - \alpha_\text{lead}

After the maneuver, obviously, the phase angle is zero.

If the interceptor and target are not in the correct configuration, the interceptor will have to wait for the next launch window. Say the phase angle at this moment is $\phi$ . The wait time $T_w$ is given by

\tag{3.6.12} 2n\pi + \phi_0 - \phi = (\omega_\text{target} - \omega_\text{interceptor}) \cdot T_w

where $n$ is a non-negative integer.

Bi-elliptic Transfer

The Bi-elliptic transfer is an orbital maneuver that moves a spacecraft from one orbit to another and may, in certain situations, require less delta-v than a Hohmann transfer maneuver. The bi-elliptic transfer consists of two half-elliptic orbits. From the initial orbit, a first burn expends delta-v to boost the spacecraft into the first transfer orbit with an apoapsis at some point $r_b$ away from the central body. At this point a second burn sends the spacecraft into the second elliptical orbit with periapsis at the radius of the final desired orbit, where a third burn is performed, injecting the spacecraft into the desired orbit.

The bi-elliptic transfer is more efficient than the Hohmann transfer when the ratio of the final to initial orbit radius is greater than 11.94. However, it takes much longer to complete than the Hohmann transfer, and hence is not commonly used.

Gravity assists

A spacecraft when launched from Earth, has to first leave Earth’s Hill sphere. After that it enters the interplanetary space with some geocentric velocity, which together with Earth’s orbital velocity puts it into a heliocentric orbit to, say, Mars. During the transfer, it will be perturbed slightly by the Earth. As it enters Martian Hill sphere, it is said to be solar perturbed about that planet. Its subsequent solar orbit will be quite different from its pre-Martian encounter orbit.

Planetary flypasts have the ability to change the energy of small bodies, and give it an entirely different heliocentric orbit. A gravity assist is a maneuver that uses the gravitational field of a planet or moon to change the speed and direction of a spacecraft. The spacecraft approaches the planet or moon, and as it passes close to the body, it is accelerated by the gravitational field. The spacecraft then moves away from the body, having gained speed and changed direction. Gravity assists are often used to increase the speed of a spacecraft, allowing it to reach its destination more quickly or to save fuel.

Let a spacecraft having a velocity $v$ relative to the sun, approach Jupiter. From Jupiter’s frame, the spacecraft has a velocity of $v_J - v$ . The velocity after the slingshot, from Jupiter’s frame is the same, $v_J - v$ , but its direction has changed. However, from a heliocentric frame, the velocity is $v_J + (v_J - v) = 2v_J - v > v$ . The spacecraft has gained a velocity of $2(v_J - v)$ in the heliocentric frame.

The orbital elements of the spacecraft before and after the flyby can be much different. A quantity that remains approximately constant during the flyby is the Tisserand parameter, which is simply the jacobian constant of the circular restricted three-body problem when $m_1 \gg m_2$ . The Tisserand parameter is given by

\tag{3.6.13} T = \frac{R}{a} + 2 \sqrt{\frac{a}{R}(1-e^2)} \cos i

where $a$ is the semi-major axis of the spacecraft’s orbit, $e$ is the eccentricity of the spacecraft’s orbit, $i$ is the inclination of the spacecraft’s orbit, and $R$ is the semi-major axis of the orbit of the planet. The Tisserand parameter is a useful tool for understanding the dynamics of small bodies in the solar system, and for planning spacecraft trajectories that involve flybys of planets.

Close approaches between the planets and minor bodies (asteroids or comets) of Solar System are very common. As a consequence, the orbits of minor bodies can change drastically over a very short time interval. The aim of this problem will be to construct a certain function (the Tisserand’s parameter) of orbital elements, such that its value remains unchanged when the asteroid’s orbit is perturbed as a result of a close encounter with a planet. It should follow that this quantity is very useful for studying apparent discontinuities in orbital evolution of minor bodies in Solar System.

Consider a setup in which an object (a planet) of mass $m$ orbits in a circular trajectory with radius $R$ around a large central body (a star) of mass $M \gg m$ .

a) Write down an expression for the orbital speed $V$ of the planet around the central star in terms of $G, M$ and $R$ .

In addition, let us consider a small object (an asteroid) with mass $\mu \ll m \ll M$ , which orbits the central star in an elliptical trajectory with semi-major axis $a$ and eccentricity $e$ . Let us denote by $i$ the inclination of the asteroid’s orbit with respect to the plane of the planet’s orbit. For the sake of simplicity, let us first assume that $i = 0$ .

b) Write down an expression for the speed $v$ of the asteroid as it passes at the distance $R$ from the central body. Express your result in terms of $V, R$ and $a$ .

c) Find the projection $v_\parallel$ of the asteroid’s velocity in the direction tangent to the planet’s orbit. Express your result in terms of $V, R, a$ and $e$ .

Now suppose that a close approach between the asteroid and the planet takes place.

d) Calculate the speed $u$ (relative to the planet) at which the asteroid enters the planet’s sphere of gravitational influence. Express your result in terms of the quantities $V, R, a$ and $e$ .

e) Discuss what adjustments would need to be made to the result of part c) for the projection $v_\parallel$ in the case of a non-zero inclination $i$ .

f) Similarly, explain how your answer in part d) would change in the case of non-zero inclination $i$ . In particular, express $u$ in terms of $V$ and the Tisserand’s parameter

\mathcal{T} = \frac{R}{a} + 2\sqrt{\frac{a}{R}(1-e^2)}\cos i .

This we can understand as a function $\mathcal{T}(a, e, i)$ of the orbital elements $a, e$ and $i$ of the asteroid.

The elements of the asteroid’s orbit are expected to change as a consequence of the close encounter with the planet. Let us denote their new values as $a', e'$ and $i'$ .

g) Justify why the value of the Tisserand’s parameter of the asteroid’s orbit does not change after the asteroid makes a close encounter with the planet. In other words, show that

\mathcal{T}(a, e, i) = \mathcal{T}(a', e', i') .

In October 2018, the asteroid 2018 UA was observed to exhibit a sudden change in its orbital elements: the original values $a \simeq 2.873 \times 10^{11} \text{ m}$ , $e \simeq 0.5470$ , $i \simeq 6.368^\circ$ changed within a few days to new values $a' \simeq 2.080\times 10^{11} \text{ m}$ , $e' \simeq 0.4474$ , $i' \simeq 2.644^\circ$ . It is therefore natural to suspect that the asteroid underwent a close flyby either at the Earth or at Mars.

h) Calculate the orbital radius $R$ of the perturbing body. Use this result to decide whether the planet passed by the Earth or Mars. For the sake simplicity, assume that the orbital inclination of the perturbing body relative to the plane of the ecliptic was zero.

a) By equating gravitational force and centripetal force:

\frac{GMm}{R^2} = \frac{mV^2}{R} \implies \boxed{V = \sqrt{\frac{GM}{R}}}

b) From the vis-viva equation:

v^2 = GM \left( \frac{2}{R} - \frac{1}{a} \right) = \frac{GM}{R} \left( 2 - \frac{R}{a} \right) \implies v = \boxed{\sqrt{2 - \frac{R}{a}} V}

c) From the conservation of angular momentum at the intersection point $R$ :

r_p v_p = r_a v_a = R v_\parallel

where $r_p, v_p$ and $r_a, v_a$ are coordinates at periapsis and apoapsis. We know that the specific angular momentum is $h = \sqrt{GMa(1-e^2)}$ .

R v_\parallel = \sqrt{GMa(1-e^2)} = \sqrt{\frac{GM}{R} Ra(1-e^2)}

v_\parallel = \sqrt{\frac{GM}{R}} \sqrt{\frac{a}{R}(1-e^2)} = \boxed{\sqrt{\frac{a}{R}(1-e^2)} V}

d) The velocity $\vec{v}$ of the asteroid can be decomposed into a tangential component $\vec{v}_\parallel$ and a radial component $\vec{v}_\perp$ . The magnitude is $v^2 = v_\parallel^2 + v_\perp^2$ , so

v_\perp^2 = v^2 - v_\parallel^2 = \left( 2 - \frac{R}{a} - \frac{a}{R}(1-e^2) \right) V^2

The relative velocity $\vec{u} = \vec{v} - \vec{V}$ , and since $\vec{V}$ is entirely tangent to the orbit, $\vec{u} = (\vec{v}_\parallel - \vec{V}) + \vec{v}_\perp$ . Thus

u^2 = (\vec{v}_\parallel - \vec{V})^2 + v_\perp^2 = v_\parallel^2 - 2v_\parallel V + V^2 + v^2 - v_\parallel^2 = v^2 + V^2 - 2v_\parallel V

\boxed{u = \left[ 3 - \frac{R}{a} - 2\sqrt{\frac{a}{R}(1-e^2)} \right]^{1/2} V}

e) In the case of non-zero inclination, the projection $v_\parallel$ into the planet’s orbital plane (the direction of $\vec{V}$ ) must account for the angle $i$ . The new tangent component in the planet’s plane is:

\boxed{v_\parallel = \sqrt{\frac{a}{R}(1-e^2)} V \cos i}

f) In the formula for $u^2$ from part d), only $v_\parallel$ changes. So,

u^2 = v^2 + V^2 - 2v_\parallel V = V^2 \left( 2 - \frac{R}{a} \right) + V^2 - 2 \sqrt{\frac{a}{R}(1-e^2)} V^2 \cos i

u^2 = \left( 3 - \left( \frac{R}{a} + 2\sqrt{\frac{a}{R}(1-e^2)}\cos i \right) \right) V^2 = (3 - \mathcal{T}) V^2

\boxed{u = \sqrt{3 - \mathcal{T}}\, V}

g) Since the mass of the asteroid is much smaller than the mass of the planet and the central star ( $\mu \ll m \ll M$ ), the gravitational scatter process can be modeled as a two-body encounter in the rest frame of the planet. In this frame, energy is conserved, meaning the speed $u$ at which the asteroid approaches the planet is equal to the speed at which it leaves the planet’s sphere of influence ( $u = u'$ ). Since $V$ is also practically unchanged for the planet, we have:

3 - \mathcal{T} = 3 - \mathcal{T}' \implies \boxed{\mathcal{T}(a,e,i) = \mathcal{T}(a',e',i')}

h) Using the invariance of the Tisserand’s parameter, $\mathcal{T} = \mathcal{T}'$ :

\frac{R}{a} + 2\sqrt{\frac{a}{R}(1-e^2)}\cos i = \frac{R}{a'} + 2\sqrt{\frac{a'}{R}(1-e'^2)}\cos i'

R = \left\{ \frac{2a a'}{a' - a} \left[ \sqrt{a'(1-e'^2)}\cos i' - \sqrt{a(1-e^2)}\cos i \right] \right\}^{2/3}

Substituting the given values:

\boxed{R \approx 1.498 \times 10^{11} \text{ m}}

Since $1 \text{ AU} \approx 1.496 \times 10^{11} \text{ m}$ , the planet is $\boxed{\text{Earth}}$ .

Problems

Consider a spacecraft currently in a circular orbit of radius

a_1 = 1 \mathrm{\,AU}

around sun, and has to go into a circular orbit of radius

a_2 = 5 \mathrm{\,AU}

, via a bi-elliptic transfer. To achieve this, it first goes into an elliptical orbit, with its aphelion at a distance

r = 8 \mathrm{\,AU}

. Then it goes into an orbit whose perihelion is at

a_2

. Find 1.

\Delta v

of the first burn 2.

\Delta v

of the second burn 3.

\Delta v

of the third burn 4. minimum fuel needed for the maneuver, if the exhaust velocity is

v_\text{exh} = 1 \mathrm{\,km/s}

and mass of the empty spacecraft is

m_0 = 2000 \mathrm{\, kg}

. Is this maneuver feasible?

The semi-major axis of the first parking orbit will be

a_\text{p1} = \frac{a_1 + r}{2} = 4.5 \mathrm{\,AU}

The velocity at perihelion ( $a_1$ ) can be found via the vis-viva equation

v_\text{p1}^2 = \mu \left( \frac{2}{a_1} - \frac{1}{a_\text{p1}} \right) \implies v_\text{p1} \approx 39.7 \mathrm{\,km/s}

The speed of the spacecraft in the circular orbit at $a_1$ was

v_1 = \sqrt{\frac{\mu}{a_1}} \approx 29.8 \mathrm{\,km/s}

Hence the Delta-v for the first burn is

\Delta v = v_\text{p1} - v_1 = \boxed{ 9.9 \mathrm{\,km/s} }

The semi-major axis of the second parking orbit will be

a_\text{p2} = \frac{r + a_2}{2} = 6.5 \mathrm{\,AU}

The velocity at aphelion ( $r$ ) is

v_\text{a2}^2 = \mu \left( \frac{2}{r} - \frac{1}{a_\text{p2}} \right) \implies v_\text{a2} \approx 9.2 \mathrm{\,km/s}

The speed of the spacecraft at the aphelion in the first parking orbit was

v_\text{a1}^2 = \mu \left( \frac{2}{r} - \frac{1}{a_\text{p1}} \right) \implies v_\text{a1} \approx 5.0 \mathrm{\,km/s}

Hence the Delta-v for the second burn is

\Delta v = v_\text{a1} - v_\text{a2} = \boxed{ 4.2 \mathrm{\,km/s} }

The velocity at perihelion of the second parking orbit ( $a_2$ ) is

v_\text{p2}^2 = \mu \left( \frac{2}{a_2} - \frac{1}{a_\text{p2}} \right) \implies v_\text{p2} \approx 14.8 \mathrm{\,km/s}

The speed of the spacecraft in the circular orbit at $a_2$ will be

v_2 = \sqrt{\frac{\mu}{a_2}} \approx 13.3 \mathrm{\,km/s}

Hence the Delta-v for the third burn is

\Delta v = v_\text{p2} - v_2 = \boxed{ 1.5 \mathrm{\,km/s} }

The total Delta-v of the maneuver is given by

\Delta v = \Delta v_1 + \Delta v_2 + \Delta v_3 = 15.6 \mathrm{\,km/s}

To find the minimum fuel, we take the fuel left after the maneuver to be zero. By Tsiolkovsky’s equation,

\Delta v = v_\text{exh} \ln \left( \frac{m_0 + m_\text{fuel}}{m_0} \right)

\implies m_\text{fuel} = m_0 \left[ \exp \left( \frac{\Delta v}{v_\text{exh}}\right) -1 \right] \approx \boxed{ 1.2 \times 10^{10} \mathrm{\,kg} }

By no means is this maneuver feasible.

Three Body Systems Advanced Topics