Where in orbit should a rocket fire to escape with minimum ΔV

In summary: What I would do then is to minimise the variable ##\Delta v## in terms of the variable ##v##.You can, of course, assume that it must be at the apogee or perigree and consider only two speeds. But, you might as well prove this as...Proving that it must be at the apogee or perigree is straightforward: at one of those two points, the rocket is at its highest potential energy, so minimising ##\Delta v## there will give the best result.
  • #1
radio360
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1
Homework Statement
A rocket is in elliptic orbit around the Earth. To put it into an
escape orbit, its engine is fired briefly, changing the rocket’s
velocity by ΔV. Where in the orbit, and in what direction, should
the firing occur to attain escape with a minimum value of ΔV
Relevant Equations
Vesc = sqrt(2Gm/r)
I am really lost on how to deal with this. Since this is an elliptic orbit, the mechanical energy is negative. For the rocket to escape orbit, we have to get the mechanical energy to be equal to or greater than zero. I thought at first that it would escape in the perigee, since that's where the rocket's velocity is fastest, so it would need the minimum nudge ΔV to make it escape. However, the apogee also has the highest gravitational potential energy, so the rocket would require a lower velocity to escape. Either way, it is at one of the end-points, but I am not sure which and how to prove it.
 
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  • #2
Given how well you've described the nature of the problem, it's not clear why you can't go ahead and solve it.

Can you write down the energy equation?
 
  • #3
Try considering the same ΔV at both points and figure out which one changes the rocket's energy the most.
 
  • #4
PeroK said:
Given how well you've described the nature of the problem, it's not clear why you can't go ahead and solve it.

Can you write down the energy equation?
vela said:
Try considering the same ΔV at both points and figure out which one changes the rocket's energy the most.
I guess I am lost on how to express this mathematically.
This is how I went about it:
At apogee:(1/2)mV2^2-GM/r2 = 1/2m(V2+ΔV)^2
At perigee:(1/2)mV1^2-GM/r1=1/2m(V1+ΔV)^2
Where r2>r1, V1>V2
At the perigee, the rocket's energy changed the most since it was most negative and changed to a positive value, and 1/2m(v1+ΔV)^2 > 1/2m(v2+ΔV)^2. Since the radius of the gravitational potential goes to infinity, the change is largest when V is largest in the orbit, which is at the perigee.
Would this be a sufficient/mathematically correct answer?
 
  • #5
I don't quite understand what you've done. The ##\Delta v## is different in the two cases.
 
  • #6
PeroK said:
I don't quite understand what you've done. The ##\Delta v## is different in the two cases.
Ah right I got lost. That means you would need a smaller amount of Delta V (the minimum) in the perigee case than you need in the apogee case to achieve the same value 1/2m(v1+ΔV)^2 = 1/2m(v2+ΔV)^2.
 
  • #7
Another way to ponder this is to consider the propellant. Assume unchanging exhaust velocity (specific impulse). The extra potential energy it has at apogee will be lost if it is consumed there: when consumed at perigee it gets added back to the spacecraft.
 
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  • #8
radio360 said:
I guess I am lost on how to express this mathematically.
This is how I went about it:
At apogee:(1/2)mV2^2-GM/r2 = 1/2m(V2+ΔV)^2
At perigee:(1/2)mV1^2-GM/r1=1/2m(V1+ΔV)^2
Where r2>r1, V1>V2
The energy before and after the ##\Delta v## aren't going to be the same, so you shouldn't be setting anything equal to each other.

At the perigee, the rocket's energy changed the most since it was most negative and changed to a positive value, and 1/2m(v1+ΔV)^2 > 1/2m(v2+ΔV)^2.
It doesn't change to a positive value. I'm not sure why you ignored the potential energy after the change in velocity.

You want to show that ##\frac 12 m(v_1 + \Delta v)^2 - \frac 12 m v_1^2 > \frac 12 m(v_2 + \Delta v)^2 - \frac 12 mv_2^2##. So to achieve the same change in energy, you would need a smaller ##\Delta v## at perigee.
 
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  • #9
vela said:
The energy before and after the ##\Delta v## aren't going to be the same, so you shouldn't be setting anything equal to each other.It doesn't change to a positive value. I'm not sure why you ignored the potential energy after the change in velocity.

You want to show that ##\frac 12 m(v_1 + \Delta v)^2 - \frac 12 m v_1^2 > \frac 12 m(v_2 + \Delta v)^2 - \frac 12 mv_2^2##. So to achieve the same change in energy, you would need a smaller ##\Delta v## at perigee.
Could we say that since it fired briefly, the potential energy doesn't change in that brief time-span but the kinetic energy does at that brief moment?
 
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  • #10
radio360 said:
Could we say that since it fired briefly, the potential energy doesn't change in that brief time-span but the kinetic energy does at that brief moment?
Yes, assume the change in PE is negligible.
 
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  • #11
PeroK said:
Yes, assume the change in PE is negligible.
Thank you!!
 
  • #12
The key point is that the change in KE, hence the change in total energy, due to a sudden boost in speed is:
$$\Delta E = \frac 1 2 m(v +\Delta v)^2 - \frac 1 2 mv^2$$And, in this problem, the required ##\Delta E## is constant, as total mechanical energy is conserved throughout the orbit. So ##\Delta E## is the difference between the escape energy and energy of the orbit.

What I would do then is to minimise the variable ##\Delta v## in terms of the variable ##v##.

You can, of course, assume that it must be at the apogee or perigree and consider only two speeds. But, you might as well prove this as well.
 
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  • #13
vela said:
So to achieve the same change in energy, you would need a smaller Δv at perigee.
Initially , this seems to imply that, by firing at perigee, you get something for nothing. That can't be right and, as 'someone' pointed that out to me, at perigee, the fuel in the tank also has higher KE. It will transfer the same amount of Momentum to the craft (i.e. velocity increase) but, as KE is proportional to v2, there is also a difference in KE of the ejecta.

In slingshot transfers, they choose the perigee to the guest planet to get the most energy from the event.
 
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  • #14
sophiecentaur said:
there is also a difference in KE of the ejecta.
Indeed. Energy is conserved. If we want to maximize the fuel energy delivered to the craft, we need to minimize the energy lost to the ejecta. The change in the KE of the ejecta is minimized when the craft's forward velocity is maximized.
 
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  • #15
The problem is also asking about the direction in which the impulse is to be applied. Is that obvious? Maybe it is, but OP needs to say something about it.
 
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  • #16
I think there is more to this problem than has already been said. It is easier to see what's going on if one writes the total energy in terms of the centrifugal potential, $$E=\frac{1}{2}mv_r^2+\frac{L^2}{2mr^2}-\frac{GMm}{r}.$$ Two points are clear from this equation regarding optimization of ##\Delta v##:
  1. The impulse must be delivered in a direction that increases the first term and not the second term.
  2. The impulse must be delivered at a point where ##v_r## is at a maximum as the satellite goes around its orbit. This point is neither at perigee nor at apogee, where ##v_r=0##, but somewhere in-between.
This is still a live homework problem so I will stop here.
 
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  • #17
Unless you set a trajectory that crashes into the planet, then direction is not important. You have the same escape velocity in all directions.
 
  • #18
PeroK said:
Unless you set a trajectory that crashes into the planet, then direction is not important. You have the same escape velocity in all directions.
Yes. However, if you already have a velocity in a given direction then any delta-V should be aligned with that direction. As opposed to, for example, the opposite direction.
 

Related to Where in orbit should a rocket fire to escape with minimum ΔV

What is ΔV and why is it important for escaping orbit?

ΔV, or delta-v, represents the change in velocity required for a spacecraft to perform a specific maneuver, such as escaping Earth's orbit. It is crucial because minimizing ΔV translates to less fuel consumption, making the mission more efficient and cost-effective.

Where in orbit should a rocket fire to achieve minimum ΔV for escape?

A rocket should fire at the point in its orbit where it is moving fastest, which is typically at the periapsis (the closest point to Earth). This allows the rocket to take advantage of the Oberth effect, where burning fuel at higher velocities results in a more efficient transfer of energy.

What is the Oberth effect and how does it help in minimizing ΔV?

The Oberth effect occurs when a rocket engine generates thrust at higher velocities, such as at periapsis. The kinetic energy gained from the burn is maximized because the rocket is already moving quickly. This effect helps minimize the ΔV required to escape orbit, making the maneuver more efficient.

How does the altitude of the periapsis influence the ΔV required for escape?

The altitude of the periapsis influences the ΔV because a lower periapsis means the rocket is closer to Earth and moving faster due to gravitational acceleration. This higher velocity at a lower altitude allows for a more efficient burn, reducing the overall ΔV needed to escape Earth's gravitational pull.

Are there any specific orbital mechanics principles that govern the optimal firing point for minimum ΔV?

Yes, the principles of orbital mechanics, particularly the vis-viva equation and the conservation of angular momentum, govern the optimal firing point. These principles dictate that firing at periapsis, where the spacecraft's velocity is highest, will minimize the ΔV required for escape due to the efficient use of kinetic energy and the Oberth effect.

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