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I've been playing around with the ideal rocket equations for a week or so, just because of sheer fun. Now, my question is, how do I calculate the $\Delta v$ needed to put a ship at a certain orbit?

I have an educated guess, and the number seem to work out. Not taking atmospheric drag into account, I could do the following.

So, we know that gravitational potential energy $E_p = -\frac{GMm}{r}$. Say we are leaving Earth's surface with no initial velocity. We would need to provide sufficient $\Delta v$ such that at the target orbit, $\Delta E_c = -\Delta E_p$, because of conservation of mechanical energy.

Thus, $\Delta E_c = -\Delta E_p = \frac{GMm}{r_o} - \frac{GMm}{r_f} = \frac{1}{2}mv^2$.

So finally, this should yield $\Delta v = \sqrt{\frac{2GM}{r_o}} - \sqrt{\frac{2GM}{r_f}}$ plus any extra speed we wanted to have at that height, maybe enough to sustain a stable orbit. Is my reasoning correct?

For LEO (low earth orbit), Wikipedia claims the gravity drag would require between 1.5 - 2 extra km/s of $\Delta v$, and my math says for a 2000km high orbit, the gravity drag should be around 1.6 km/s. It sounds good, but I'd like to check.

Thank you all very much.

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  • $\begingroup$ Try using the Virial theorem to determine the change in speed necessary to change altitude (ignoring any drag effects, of course). $\endgroup$ – honeste_vivere Jun 4 '17 at 17:06
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So finally, this should yield $\Delta v = \sqrt{\frac{2GM}{r_o}} - \sqrt{\frac{2GM}{r_f}}$ plus any extra speed we wanted to have at that height, maybe enough to sustain a stable orbit. Is my reasoning correct?

No, it's not. You are calculating the difference between escape velocity at two different altitudes.

To put something into orbit, the object needs to gain kinetic and potential energy. The amount of energy expended (delta V) depends very much on the path taken to achieve this goal. Suppose we could teleport to the desired orbit, a circular orbit some height $h$ above the surface of an airless, non-rotating planet. The teleport device of course obeys conservation of energy. The initial specific energy is $E_0 = -\frac \mu r$, where $\mu \equiv GM$ is the planet's standard gravitational parameter. The final specific energy is $E_1 = \frac {v^2} 2 - \frac \mu {r+h} = -\frac 1 2 \frac \mu{r+h}$. The difference is the minimum specific energy needed to achieve that orbit: $$\Delta E = \frac\mu r-\frac 12\frac\mu {r+h} = \frac\mu2\left(\frac2r - \frac1{r+h}\right)$$ We can't teleport, of course. Any real scheme will require more energy than this. Not quite real yet, suppose the vehicle applies an impulsive (instantaneous) delta V at the surface of the planet, with the velocity vector angled at some angle $\phi$ above the horizontal and the magnitude just enough to have the object reach a height $h$ above the surface at aphelion. Then the vehicle applies another impulsive burn to circularize the orbit.

The semi-major axis, initial velocity, aphelion velocity, and eccentricity of the transfer orbit (the path from the surface to the aphelion point) are $$\begin{aligned} a &= \frac{r+h}{1+\epsilon} \\ {v_i}^2 &= \mu \left(\frac2r-\frac1a\right) \\ {v_a}^2 &= \mu \frac{1-\epsilon}{r+h} \\ \epsilon &= \frac{h^2 + r(r+2h)\sin^2\phi}{h(2r+h)+r^2\sin^2\phi} \end{aligned}$$ The first is simple geometry. The aphelion distance is $a(1+\epsilon)$, and we want that to be $r+h$. The second and third are the vis-viva equation, which comes in very handy for these kinds of calculations. The final expression is a tedious calculation involving the eccentricity vector, $\vec \epsilon = \frac {\vec v \times (\vec r \times \vec v)}\mu - \hat r$, the semi-major axis, and the initial velocity.

An example using $\mu = 398600.4418\,\text{km}^3/\text{s}^2$ (the Earth's gravitational parameter), $r = 6378.1366\,\text{km}$ (the Earth's equatorial radius), $h=2000\,\text{km}$ (the altitude in the question), and various angles follows.

phi(°)   ecc     a(km)   v1(km/s)  v2(km/s)  dv2(km/s) dvtot(km/s) loss(km/s)
   0   0.135536 7378.137 8.424077  6.413111  0.484444    8.908522   0.110036
  30   0.517815 5519.867 7.264824  4.789631  2.107924    9.372747   0.574261
  60   0.893738 4424.125 5.906997  2.248450  4.649105   10.556103   1.757617
  90   1.0      4189.068 5.462334  0.0       6.897555   12.359889   3.561403

In the above, phi is the angle above horizontal, ecc is the eccentricity of the transfer orbit, a is the semi-major axis of the transfer orbit, v1 is the initial velocity needed to achieve that transfer orbit, v2 is the velocity on the transfer orbit at aphelion (which is occurs at 2000 km above the surface of the Earth), dv2 is the delta V needed to circularize the orbit at aphelion, dvtot is the sum of v1 and dv2, and loss is amount by which dvtot exceeds the ideal teleport delta V (8.798486 km/s, in this example).

A key takeaway point: On an airless planet (or moon), it makes much more sense to launch horizontally than vertically. Doing so from the surface of the Earth makes no sense. This would mean flying at Mach 25 through the thickest part of the atmosphere. Rockets launch vertically because the first goal is to get above the thickest part of the atmosphere. Rockets then start a turn toward the horizontal so as to avoid an overly huge penalty for flying vertically.

I used a very simple python script to generate the above table. Python is a handy language for such analyses. Here's the script:

import math

mu = 398600.4418   # Earth gravitational parameter, in km^3/s^2
r  = 6378.1366     # Earth equatorial radius, in km (IERS 2003)
h  = 2000.0        # Desired altitude, in km

vc = math.sqrt(mu/(r+h))
dv_opt = math.sqrt(mu*(2.0/r - 1.0/(r+h)))
print(dv_opt)

for f in range(0,91,30):
    theta = f * math.pi/180.0
    sintsq = math.sin(theta)**2
    ecc = (h*h+(r*r+2*r*h)*sintsq)/(h*h+2*r*h+r*r*sintsq)
    a = (r+h)/(1.0+ecc)
    v1 = math.sqrt(mu*(2.0/r-1.0/a))
    v2 = math.sqrt(mu*(2.0/(r+h)-1.0/a))
    dv = v1 + (vc - v2)
    print("{:2d} {:.6f} {:8.6f} {:8.6f} {:8.6f} {:8.6f} {:9.6f} {:8.6f}"
        .format(f, ecc, a, v1, v2, vc-v2, dv, dv-dv_opt))
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  • $\begingroup$ I have yet much to learn in the realm of rocket science, I can see from this post. However, thank you very much for providing me an insight as to why I was wrong. $\endgroup$ – ChemiCalChems Jun 5 '17 at 6:44
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That extra speed, enough to sustain a stable orbit, makes all the difference! We need to compute the speed on a circular orbit, to keep things simple. The radial acceleration is equal to the pull of gravity:

$$ m\frac{\Delta v}{r_f} = \frac{GMm}{r_f^2},$$

which results in the ultra-classic

$$ \Delta v = \sqrt{\frac{GM}{r_f}}.$$

But now you deal with a rocket here and there is an essential point your forgot: the rocket will loose mass as it burns fuel! Let's then analyse the scheme you developed in your question and in your comments to this answer of mine. You want to do two steps.

  • In step 1, your rocket, fully loaded with a mass $m_0$, starts at radius $r_o$ with a speed of 0. It then reaches radius $r_f$ with some speed $v_1$, and its mass has been reduced to $m_1 < m_0$.

  • In step 2, your rockets then reaches the same altitude $r_f$ but at the orbital speed $v_2$ I wrote above, with a further reduced mass $m_2 < m_1$.

Let us compute the energy balance for each step. We need to take into account the energy produced by the rocket engine. Let's say that burning a mass $m$ of fuel produces an energy $km$ for some constant $k$. The value of $k$ could be computed from the enthalpy of the combustion reaction.

In step 1,

$$ k (m_0 - m_1) = \left(\frac{1}{2}m_1 v_1^2 - \frac{GMm_1}{r_f}\right) - \left(0 - \frac{GMm_0}{r_0}\right)$$

In step 2,

$$k(m_1 - m_2) = \left(\frac{1}{2}m_2 v_2^2 - \frac{GMm_2}{r_f}\right) - \left(\frac{1}{2}m_1 v_1^2 - \frac{GMm_1}{r_f}\right)$$

By summing these two equations

$$k(m_0 - m_2) = \left(\frac{1}{2}m_2 v_2^2 - \frac{GMm_2}{r_f}\right)- \left(0 - \frac{GMm_0}{r_0}\right)$$

which is the amount of fuel to burn to do it in one step, let's call it D for "direct", from motionless at radius $r_o$ to orbital speed at $r_f$. No surprise here, we are dealing with conservative forces but I just wanted to make explicit to you.

There is a big problem however. The design of the rocket engine will fix the rate of mass decrease. Since step 1 + 2 will take longer than step D, if you have enough fuel for step D, then that same amount of fuel will be burnt out before you can finish step 1 + 2. So you will have to provision more fuel for step 1 + 2. A quantitative answer would require solving the equations of motion. So clearly, just looking at energy balances is not enough here.

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  • $\begingroup$ I do understand. However, would what the energy be to reach a certain height, and then fall? Wouldn't it be of the form $-\Delta E_p = \Delta E_c$? Then you could add the kinetic energy to sustain orbit if you'd ever like to, of course but that's not completely my question. $\endgroup$ – ChemiCalChems Jun 4 '17 at 21:03
  • $\begingroup$ I guess what I'm trying to say is I want to get to a certain without necessarily having the mechanical energy to sustain an orbit. Maybe I'd want more speed than necessary, or maybe less. $\endgroup$ – ChemiCalChems Jun 4 '17 at 21:05
  • $\begingroup$ Yes, what you computed is indeed the energy to reach a certain height, and then fall back. Ok, sorry then, I misunderstood your question. $\endgroup$ – user154997 Jun 4 '17 at 21:06
  • $\begingroup$ However, on reaching the height, if I accelerated to orbital speed, then I would sustain an orbit, right? So going to orbit directly and doing the process in 2 steps would be equivalent. $\endgroup$ – ChemiCalChems Jun 4 '17 at 21:08
  • $\begingroup$ I added to my answer. $\endgroup$ – user154997 Jun 5 '17 at 8:23

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