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I think this is a very very simple question for me, but unfortunately I cant solve it...

A satellite is orbiting in a circular path at an altitude of 600 kilometers. If the satellite is capable of providing a total velocity increment (ΔV) of 1.2 kilometers per second, what would be the maximum possible circular orbit that the satellite can reach?

I try to use the equation I found on "fundamentals of astrodynamics and applications" this book.
And assume it is a circle orbit, than caculate it with matlab.
however the result value I got is negative, which not correct.
Can anyone tell me what did I do wrong? thank you!!

My code:

clc;
clear;

mu = 398600.4415; % Standard gravity parameters,km^3/s^2
earth_radius = 6378.137; % Earth radius,km
initial_altitude = 600; % initial orbit altitude,km
deltaV = 1.2; % deltaV,km/s

% initial orbit altitude
ri = earth_radius + initial_altitude;

% initial speed
Vi = sqrt(mu / ri);

% final initial
Vf = Vi + deltaV;

% final orbit altitude
rf = mu / Vf^2;

% final orbit altitude
final_altitude = rf - earth_radius;

fprintf('final orbit altitude:%f km\n', final_altitude);

final orbit altitude:-1181.274068 km

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    $\begingroup$ Don't forget about potential energy. en.wikipedia.org/wiki/Hohmann_transfer_orbit $\endgroup$
    – PM 2Ring
    Nov 11, 2023 at 10:10
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    $\begingroup$ Your issue is that in a larger orbit, the craft will have a slower velocity. A better way to solve would be to compute the delta V for a Hohmann transfer which includes 2 burns, one for creating the elliptical transfer orbit and another for circularizing at the apex of the transfer. But @PM2Ring is correct in that you should probably take an energy based approach to the problem, as a Hohmann transfer is not always the most efficient means of orbit raising and may not give the “highest possible orbit”. But finding the most optimal transfer is a very complex task,so go with an energy balance $\endgroup$
    – A McKelvy
    Nov 11, 2023 at 15:10
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    $\begingroup$ For a moderate available delta-v of 1200 m/s compared to the initial circular orbital velocity of $\sqrt{GM/a}$ of 7558 m/s, I think you should at least start learning the math by trying the two impulse approach. Use the vis-viva equation which is I used to get your initial velocity using GM =3.986E+14 m/s and $r=a$ (radius = semimajor axis) and $r=r_0 + 600,000$ m and $r_0 = 6378137$ m (equatorial radius of Earth, the convention when quoting orbital altitudes). $\endgroup$
    – uhoh
    Nov 11, 2023 at 16:26
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    $\begingroup$ @uhoh I am currently writing an answer where I will show the math and how to calculate it. I think I will be done in about 30 minutes, since it is hard to write formulas in SE $\endgroup$ Nov 11, 2023 at 16:28
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    $\begingroup$ Your first delta-v boosts $a$, you are now at periapsis, so you can calculate apoapsis. Your second delta-v boosts $a$ again to make it equal to your apoapsis $r$. So your job is to find pairs of delta-v that add up to 1200 but end with $a=r$. $\endgroup$
    – uhoh
    Nov 11, 2023 at 16:29

5 Answers 5

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To calculate how high your satellite will go, you forgot 1 major thing. The hohmann transfer. I am going to show you how to calculate the transfer from a 600 km circular orbit to another circular orbit.

We can divide it into 3 steps. The first step is to calculate how fast the satellite is going at 600 km. The second step is to calculate the speed it will have after conducting a hohmann transfer and the third step is to calculate how much faster it will have to be going in the new circular orbit.

enter image description here

I am going to use these numbers that you have given: $$Radius\ of\ Earth= 6378.137km = 6\ 378\ 137m$$ $$Mass\ of\ Earth= 5.972\cdot\ 10^{24}kg$$ $$Gravitational\ constant= 10\cdot\ 6.673^{-11}$$

Here is more information about the gravitational constant

To calculate the velocity of a circular orbit we need to use this formula:

$$\Delta v= \sqrt{\frac{GM}{r}} $$

This formula can only be used if there is a circular orbit because it assumes that the centrifugal force equals the gravitational force.

$$F_c= F_g $$ $$\frac{m \cdot\ v^2}{r} = \frac{G\cdot\ M\cdot\ m}{r^2} $$ $$\Delta v= \sqrt{\frac{GM}{r}} $$

Where $\Delta v$ is the velocity, $M$ is the mass of Earth, $G$ is the gravitational constant and $r$ is the radius of the orbit.

If we want to know the velocity of a spacecraft with a circular orbit at a height of 600 km, then we can do the following:

$$r = radius\ of\ Earth + height\ of\ spacecraft$$ $$r = 6378.137km + 600km$$ $$r = 6978.137km = 6\ 978\ 137m$$

Now putting in the numbers in the formula:

$$\Delta v= \sqrt{\frac{GM}{r}} = \sqrt{\frac{( 6.673\cdot\ 10^{-11}) \cdot\ (5.972\cdot\ 10^{24}kg)}{6\ 978\ 137m}}= 7557.0225m/s$$

You said that that your spacecraft has 1.2km/s which would be 1200 m/s. To calculate it, I would like to divide the velocity needed into 4 different ones. enter image description here

  • v1 = the velocity I just calculated (7557.0225m/s)

  • v2 = the velocity the spacecraft will have after completing the hohmann transfer burn

  • v3 = the velocity the spacecraft will have one it arrived at apoapsis

  • v4 = the velocity the spacecraft will have once it is in its new circular orbit

As you already mentioned, the total $\Delta v$ the spacecraft has is 1200m/s.

$$\Delta v_{total} = (\Delta v_2 - \Delta v_1) + (\Delta v_4 - \Delta v_3)$$

To calculate $\Delta v_2$ and $\Delta v_3$, one needs to use a different formula to calculate it, since the spacecraft isn't in a circular orbit. The best way to do so is to use the mechanical energy.

$$mechanical\ energy = kinetic\ energy + potential\ energy$$ $$ME = KE + PE$$

Kinetic energy

$$\Delta v= \sqrt{\frac{GM}{r}} \rightarrow v^2 = \frac{GM}{r}$$ $$KE = \frac{1}{2}\cdot\ m\cdot\ v^2 = \frac{1}{2}\cdot\ m\cdot\ \frac{GM}{r}$$

Potential energy

$$PE= -\frac{G \cdot\ M \cdot\ m }{r}$$ For more information about this formula, check this video

Calculating the total energy

$$ME = KE + PE$$ $$ME = (\frac{1}{2}\cdot\ m\cdot\ \frac{GM}{r}) + (-\frac{G \cdot\ M \cdot\ m }{r})$$ $$ME = \frac{G \cdot\ M \cdot\ m}{2 \cdot\ r} + (-\frac{G \cdot\ M \cdot\ m }{r})$$

Using this, we can calculate the total energy in a circular and elliptical orbit.

Circular orbit: $$ME = -\frac{G \cdot\ M \cdot\ m}{2 \cdot\ r}$$

Elliptical orbit: $$ME = -\frac{G \cdot\ M \cdot\ m}{2 \cdot\ a}$$

What you might notice is that there is an "a" instead of "r" there and that is because "a" should be the semi-major axis of an orbit. You can find out more about the semi-major axis and the major axis here.

enter image description here

I tried to draw what it is in the program paint to make it easier to understand. The big circle with the letter "M" is meant to be a big massive object.

So you could also write it like this: $$2 \cdot\ a = r1 + r2$$ $$ME = -\frac{G \cdot\ M \cdot\ m}{r1 + r2}$$

The next step is to calculate the velocity of each point. To do so, one should divide the total energy of the elliptical orbit in kinetic and potential energy and use the kinetic energy to calculate the velocity at the apoapsis and periapsis.

$$ME = KE + PE$$ $$ -\frac{G \cdot\ M \cdot\ m}{r1 + r2} = \frac{1}{2}\cdot\ m\cdot\ v^2 + \frac{-G \cdot\ M \cdot\ m }{r}$$ $$ \frac{1}{2} \cdot\ v^2 = \frac{G \cdot\ M}{r} - \frac{G \cdot\ M}{r1 + r2}$$

$$\Delta v_{periapsis}= \sqrt{2 \cdot\ G \cdot\ M \cdot\ (\frac{1}{r1}-\frac{1}{r1+r2})}$$ $$\Delta v_{apoapsis}= \sqrt{2 \cdot\ G \cdot\ M \cdot\ (\frac{1}{r2}-\frac{1}{r1+r2})}$$

Now we have the formulas needed to calculate the speed at the periapsis and apoapsis in an elliptical orbit, so we can calculate v2 and v3.

$$ \Delta v_1 = 7557.0225m/s $$

$$\Delta v_2= \sqrt{2 \cdot\ G \cdot\ M \cdot\ (\frac{1}{r1}-\frac{1}{r1+r2})}= \sqrt{2 \cdot\ 10\cdot\ 6.673^{-11} \cdot\ 5.972\cdot\ 10^{24}kg \cdot\ (\frac{1}{6 \ 378\ 137m + 600\ 000m}-\frac{1}{(6 \ 378\ 137m + 600\ 000m)+r2})}= \sqrt{2 \cdot\ 10\cdot\ 6.673^{-11} \cdot\ 5.972\cdot\ 10^{24}kg \cdot\ (\frac{1}{6 \ 978\ 137m}-\frac{1}{(6 \ 978\ 137m)+r2})}$$

$$\Delta v_3= \sqrt{2 \cdot\ G \cdot\ M \cdot\ (\frac{1}{r2}-\frac{1}{r1+r2})}= \sqrt{2 \cdot\ 10\cdot\ 6.673^{-11} \cdot\ 5.972\cdot\ 10^{24}kg \cdot\ (\frac{1}{r2}-\frac{1}{(6 \ 378\ 137m + 600\ 000m)+r2})}= \sqrt{2 \cdot\ 10\cdot\ 6.673^{-11} \cdot\ 5.972\cdot\ 10^{24}kg \cdot\ (\frac{1}{r2}-\frac{1}{(6 \ 978\ 137m)+r2})}$$

$$\Delta v_4 = \sqrt{\frac{GM}{r2}} = \sqrt{\frac{10\cdot\ 6.673^{-11} \cdot\ 5.972\cdot\ 10^{24}kg}{r2}}$$

As you might notice, r2 is missing and that is what we are looking for. Earlier I mentioned this formula $$\Delta v_{total} = (\Delta v_2 - \Delta v_1) + (\Delta v_4 - \Delta v_3)$$ So to calculate r2, we need to place each formula for each "v" into the total "v" formula to get r2.

$$\Delta v_{total} = (\Delta v_2 - \Delta v_1) + (\Delta v_4 - \Delta v_3)$$ $$1200m/s = ( \sqrt{2 \cdot\ 10\cdot\ 6.673^{-11} \cdot\ 5.972\cdot\ 10^{24}kg \cdot\ (\frac{1}{6 \ 978\ 137m}-\frac{1}{(6 \ 978\ 137m)+r2})} - 7557.0225m/s) + (\sqrt{\frac{10\cdot\ 6.673^{-11} \cdot\ 5.972\cdot\ 10^{24}kg}{r2}} - \sqrt{2 \cdot\ 10\cdot\ 6.673^{-11} \cdot\ 5.972\cdot\ 10^{24}kg \cdot\ (\frac{1}{r2}-\frac{1}{(6 \ 978\ 137m)+r2})})$$

We can now simplify this equation to: $$\frac{V_{total} + \sqrt{\frac{G \cdot\ M}{r_1}}}{\sqrt{2 \cdot\ G \cdot\ M}} = \sqrt{\frac{1}{r1}-\frac{1}{r1+r2}}+\sqrt{\frac{1}{r2}-\frac{1}{r1+r2}}-\sqrt{\frac{1}{r2 \cdot\ 2}}$$

If you solve this equation, then r2 will be the radius you are looking for. However, my calculator was not working and after trying to fix it for 3 hours, I will leave this task for another time.

In case you want to try to calculate it, here are the values:

$$V_{total} = 1200 m/s$$ $$G = 6.673 \cdot\ 10^-11$$ $$M = 5.972 \cdot\ 10^-24kg$$ $$r_1 = 6 \ 978 \ 123m$$

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  • $\begingroup$ In the final equation, r2 is calculable since there is 1 unknown and 1 equation. My calculator was giving me a really bad time, so I decided to take a break. Anybody who has a better calculator, which is not like my 0 IQ calculator can try to give it a shot. Feel free to edit my answer! 😉 $\endgroup$ Nov 11, 2023 at 18:45
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    $\begingroup$ Never use G*M for the Sun or any of the planets. Use $\mu$ (mu) instead; the value for $\mu$ is more precise. The value in the OP is correct. In addition, there's no reason to switch from kilometers to meters. $\endgroup$ Nov 11, 2023 at 19:24
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    $\begingroup$ Note well: I am not the downvoter. You put in a lot of effort; this is not worthy of a downvote. $\endgroup$ Nov 11, 2023 at 20:17
  • $\begingroup$ @DavidHammen thanks for the tip. I will use μ from now on. $\endgroup$ Nov 11, 2023 at 20:24
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    $\begingroup$ @ Dan Mašek I edited the answer and replaced the commas with points $\endgroup$ Nov 12, 2023 at 15:05
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A single application of all of the allotted delta V will result in an elliptical orbit with a perigee equal to the original altitude. What you want to do instead is to apply a delta V at the start and then apply some more delta V half an orbit later to circularize the orbit.

Given a target radius $r_2$ for the apogee after the initial burn, the $\Delta v_1$ for that initial burn is given by $$\Delta v_1 = \sqrt{\frac {\mu}{r_i}} \left(\sqrt{\frac {2 r_2}{r_i + r_2}} - 1\right)\tag{1}$$ The $\Delta v$ needed to circularize the orbit at this new altitude is given by $$\Delta v_2 = \sqrt{\frac {\mu}{r_2}} \left(1-\sqrt{\frac {2 r_i}{r_i + r_2}}\right)\tag{2}$$ Both of the above can readily be derived from the vis-viva equation, $$v^2 = \mu\left(\frac 2 r - \frac 1 a\right)\tag{3}$$ where $v$ is the current orbital velocity, $\mu$ is the gravitational parameter for the central object (the Earth in this case), $a$ is the semi-major axis of the orbit, and $r$ is the current orbital radius. The vis-viva equation can in turn be derived from conservation of energy (kinetic plus potential) and conservation of angular momentum. However, the vis-viva equation is so very useful and so easy to memorize that it is worthwhile to commit to memory.

Aside #1: The wikipedia pages on the vis-viva equation (equation 3) and the Hohmann transfer (equations 1 and 2) are well-written and are correct. I suggest reading them as a starting point.

The sum of equations 1 and 2 yields the total $\Delta v$ needed to change from one circular orbit to another, assuming no plane change. However, you want to find what altitude can be gained given a total $\Delta v$.

A very simple way to find this is to use bisection. You know that no change in orbital radius requires zero $\Delta v$. Keep adding a positive quantity to the final orbital radius until the sum of equations 1 and 2 exceeds the target $\Delta v$. You now have the solution bracketed; the goal of bisection search is to repetitively decrease the size of the bracket by using the average of the lower and upper bounds. Solve for the $\Delta v$ at the midpoint between the previous $\Delta h$ (which yields a total $\Delta v$ less than the allotted $\Delta v$) and the current $\Delta h$ (which yields a total $\Delta v$ greater than the allotted $\Delta v$). If the total $\Delta v$ for this new $\Delta h$ is less than the target $\Delta v$,this midpoint becomes your new lower bound. If on the other hand, the total $\Delta v$ for this new $\Delta h$ is greater than the target $\Delta v$,this midpoint becomes your new upper bound. Either way you have cut the difference between the upper bound and lower bound in half.

Aside #2: There are much more advanced search techniques than bisection, but bisection is so very easy to code that it's often worth shot. In this case, it works quite nicely.

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  • $\begingroup$ Thanks for helping!! I didnt noticed if I just use all the delta V in very short time it will only change its ecc. $\endgroup$ Nov 12, 2023 at 14:28
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In addition to the great answers you have already got, beware of what happens in the extreme case!

The cost of a Hohmann transfer increases by altitude, but only up to a point. After that, the cost decreases by altitude, reaching $\Delta v = \sqrt{2} - 1$ times the original circular velocity at the limit. (The same as the escape cost, naturally). Another way to think of it is the degenerate case of a bi-elliptic transfer

So if you have a delta-v that can bring you more than 3.3 times farther out than the initial orbit, you also have enough delta-v to reach a circular orbit "at infinity" (or sufficiently far out). You can't necessarily reach all of the circular orbits in between though.

Hohmann cost

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Matlab's symbolic toolbox is very helpful to solve this problem. Below, I wrote expressions for $\Delta v$ from initial circular orbit to Hohmann transfer orbit and from Hohmann transfer orbit to final circular orbit. These expressions are functions of the radius of the final orbit, $r_2$. I can then sum these expressions to get total $\Delta v(r_2)$, set it equal to $\Delta v$ available, and solve for $r_2$.

clc
clear
close all

% mu = 3.986e5;               km^3/s^2
% R = 6378;                   km
% r1 = R + 600;               km
% dv = 1.2;                   km/s

syms mu r_1 r_2 dv

% orbit 1: eccentricity e = 0
v1 = sqrt(mu/r_1)

% orbit 2: 0<e<1
a = (r_1 + r_2)/2;          % semi-major axis
v2p = sqrt(mu*(2/r_1-1/a))  % velocity at perigee from vis-viva equation
v2a = sqrt(mu*(2/r_2-1/a))  % velocity at apogee from vis-viva equation

% orbit 3: e = 0
v3 = sqrt(mu/r_2)

% delta v from initial circular orbit to elliptical transfer orbit
dv21 = v2p-v1

% delta v from elliptical transfer orbit to final circular orbit
dv32 = v3 - v2a

% total dv is the sum
sym_eqn = dv == dv21 + dv32

% substitute in known values of mu, r1, and dv
eqn = subs(sym_eqn,[mu r_1 dv],[3.986e5 6378+600 1.2])

% solve for r2 in km (have to solve this numerically so use vpasolve instead of solve)
sol = vpasolve(eqn)
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  • $\begingroup$ Thank you for the helping!! $\endgroup$ Nov 13, 2023 at 2:09
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I used my Hohmann Spreadsheet inputting 600 km into periapsis for earth orbit and adjusting by trial and error the apoapsis for earth orbit. I look at the sum of the apoapsis and periapsis circularization burns (right side)

I get a new orbit altitude of 3511.6 kilometers

enter image description here

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