Orbital Maneuvers

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 are this book and this thing.

Delta-v

Delta-v (Δv\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 abscence of external forces, the Delta-v is equal to the change in velocity of the spacecraft.

Δv=t0t1v˙dt(3.6.1)\tag{3.6.1} \Delta v = \int_{t_0}^{t_1} |\dot{v}| dt

where v˙\dot{v} is the acceleration. When the thrust force is applied in a constant direction, Delta-v is given by Δv=v1v0\Delta v = |v_1 - v_0| 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 will be

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

where

  • vexhv_\text{exh} s the velocity of the exhaust gas in rocket frame
  • ρ\rho is the mass flow rate of the propellant

If the mass of the satellite at t=t0t=t_0 before the maneuver is m0m_0, and at t=t1t=t_1 after the maneuver is m1m_1, then the Delta-v is given by the Tsiolkovsky rocket equation

Δv=vexhln(m0m1)(3.6.3)\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 m1m0m_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 Tranfer 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).

image/svg+xml Initial Orbit 1st Burn 2nd Burn Transfer Orbit Final Orbit
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 a1a_1 to orbit 2, having radius a2a_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

at=a1+a22(3.6.4)\tag{3.6.4} a_t = \frac{a_1 + a_2}{2}

The eccentricity of the transfer orbit is

et=a2a1a2+a1(3.6.5)\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

Tt=πat3μ(3.6.6)\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, Tt=(a1+a2)332MT_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

Δv1=vp,tv1=μat(1+e1e)μa1(3.6.7)\tag{3.6.7} \Delta v_1 = v_\text{p,t} - v_1 = \sqrt{\frac{\mu}{a_t} \left( \frac{1 + e}{1 - e} \right) } - \sqrt{\frac{\mu}{a_1}}

The change in velocity at apoapsis is given by

Δv2=v2va,t=μa2μat(1e1+e)(3.6.8)\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}{1 + e} \right) }

The total Delta-v for the maneuver is

Δv=μat(1et2)(2et+1et1+et)(3.6.9)\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 a interceptor satellite which has to be maneuvered to rendezvous with a target satellite. The interceptor is in a circular orbit of radius a1a_1, and the target is in a circular orbit of radius a2a_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 αlead\alpha_\text{lead} is the angle between the initial position of the target and the (future) rendezvous point.

αlead=ωtargetTt(3.6.10)\tag{3.6.10} \alpha_\text{lead} = \omega_\text{target} \cdot T_t

where ωtarget\omega_\text{target} is the angular velocity of the target, and TtT_t is the time it takes for the interceptor to reach the target orbit. The phase angle ϕ\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

ϕ0=παlead(3.6.11)\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 ϕ\phi. The wait time TwT_w is given by

2nπ+ϕ0ϕ=(ωtargetωinterceptor)Tw(3.6.12)\tag{3.6.12} 2n\pi + \phi_0 - \phi = (\omega_\text{target} - \omega_\text{interceptor}) \cdot T_w

where nn 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 rbr_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.

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

Problems

Consider a spacecraft currently in a circular orbit of radius a1=1AUa_1 = 1 \mathrm{\,AU} around sun, and has to go into a circular orbit of radius a2=5AUa_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=8AUr = 8 \mathrm{\,AU}. Then it goes into an orbit whose perihelion is at a2a_2. Find

  1. Δv\Delta v of the first burn
  2. Δv\Delta v of the second burn
  3. Δv\Delta v of the third burn
  4. minimum fuel needed for the maneuver, if the exhaust velocity is vexh=1km/sv_\text{exh} = 1 \mathrm{\,km/s} and mass of the empty spacecraft is m0=2000kgm_0 = 2000 \mathrm{\, kg}. Is this maneuver feasible?
  1. The semi-major axis of the first parking orbit will be
ap1=a1+r2=4.5AU a_\text{p1} = \frac{a_1 + r}{2} = 4.5 \mathrm{\,AU}

The velocity at perihelion (a1a_1) can be found via the vis-viva equation

vp12=μ(2a11ap1)    vp139.7km/s 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 a1a_1 was

v1=μa129.8km/s v_1 = \sqrt{\frac{\mu}{a_1}} \approx 29.8 \mathrm{\,km/s}

Hence the Delta-v for the first burn is

Δv=vp1v1=9.9km/s \Delta v = v_\text{p1} - v_1 = \boxed{ 9.9 \mathrm{\,km/s} }
  1. The semi-major axis of the second parking orbit will be
ap2=r+a22=6.5AU a_\text{p2} = \frac{r + a_2}{2} = 6.5 \mathrm{\,AU}

The velocity at aphelion (rr) is

va22=μ(2r1ap2)    va29.2km/s 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

va12=μ(2r1ap1)    va15.0km/s 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

Δv=va1va2=4.2km/s \Delta v = v_\text{a1} - v_\text{a2} = \boxed{ 4.2 \mathrm{\,km/s} }
  1. The velocity at perihelion of the second parking orbit (a2a_2) is
vp22=μ(2a21ap2)    vp214.8km/s 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 a2a_2 will be

v2=μa213.3km/s v_2 = \sqrt{\frac{\mu}{a_2}} \approx 13.3 \mathrm{\,km/s}

Hence the Delta-v for the third burn is

Δv=vp2v2=1.5km/s \Delta v = v_\text{p2} - v_2 = \boxed{ 1.5 \mathrm{\,km/s} }
  1. The total Delta-v of the maneuver is given by
Δv=Δv1+Δv2+Δv3=15.6km/s \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,

Δv=vexhln(m0+mfuelm0) \Delta v = v_\text{exh} \ln \left( \frac{m_0 + m_\text{fuel}}{m_0} \right)     mfuel=m0[exp(Δvvexh)1]1.2×1010kg \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.