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Would a rocket that wanted to reach an elliptical orbit with a perigee of 100 miles (160 km) and an apogee of 300 miles (480 km) burn the same amount of fuel as a rocket that wanted to attain a circular orbit at 200 miles (320 km) altitude?

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  • $\begingroup$ You may find this helpful: en.wikipedia.org/wiki/Specific_orbital_energy $\endgroup$
    – PM 2Ring
    Jul 22, 2023 at 21:15
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    $\begingroup$ Pretty sure the answer is "no." Burning the fuel from x mass to y mass is directly tied to delta-V, and delta-V is tied to changes in spacecraft velocity and momentum, rather than orbital energy. Burning to a very elliptical orbit with your same periapsis is going to take less delta-v than a bi-elliptic transfer to a high circular orbit with the same semi-major axis. $\endgroup$
    – notovny
    Jul 23, 2023 at 0:00
  • $\begingroup$ @notovny I ain't talking about the same periapsis, but about the same mean value which is 200 mi in a 100 mi x 300 mi orbit. Would it take less delta-v to reach a 100x300 orbit than reaching a 200 mi circular orbit? $\endgroup$
    – Johannes
    Jul 23, 2023 at 5:28

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There is no reason to expect the required fuel to be the same.

To show this, we'll produce a simple counterexample with numbers chosen for ease of calculation. We will assume that by "same mean altitude" , OP assumes a perfectly spherical planet, and that "same mean altitude" translates directly to same semi-major axis.

We'll also use generic Distance Units ($\text{D}$) and Time units ($\text{T}$); the equations would work with any consistent unit system. In the image, assume all orbits to be going counterclockwise.

enter image description here

The central body has Standard Gravitational Parameter $\mu = 1.00\,\text{D}^3/\text{T}^2$

The spacecraft in question is in the circular Parking Orbit O (black), which has a radius of $1\,\text{D}$

We are comparing the delta-V required to go from Orbit O to Orbit C, a coplanar circular orbit with radius $3.00\,\text{D}$ (semi-major axis $3.00\,\text{D}$) and from orbit O to Orbit E, a coplanar elliptical orbit with semi-major axis $3.00\,\text{D}$ but with periapsis distance $1.00\,\text{D}$ and apoapsis $5.00\,\text{D}$

Orbit O to Orbit C:

In this example the radii of the two orbits are close enough in size that the most efficient transfer from Orbit O to Orbit C is a two impulse Hohmann transfer Maneuver;

We fire the engines at the red dot, in the direction of the red arrow to put the spacecraft into a transfer orbit (red dotted orbit), then fire the engines again at the blue dot in the direction of the blue arrow to circularize into orbit C.

Impulse 1 $$\Delta v_{1}=\sqrt{\frac{μ}{r_{1}}} \left(\sqrt{\frac{2r_{2}}{r_{1} + r_{2}}} - 1\right)$$
Impulse 2 $$\Delta v_{2}=\sqrt{\frac{μ}{r_{2}}} \left(1 -\sqrt{\frac{2r_{1}}{r_{1} + r_{2}}}\right)$$

The total delta-V required for the Hohmann transfer is the sum of the above two values.

Substitute in $r_1 = 1.00 \text{D}$ and $r_2 = 3.00\,\text{D}$ and the total delta-V for the Hohmann transfer is: $$\Delta v_{C} = \Delta v_{C1} + \Delta v_{C2} = 0.22\,\text{D/T} + 0.17\, \text{D/T} = 0.39\,\text{D/T}$$

Orbit O to Orbit E:

To calculate the delta-V requirement to go from Orbit O to Orbit E... Well, in this case, Orbit E is exactly the same as a Hohmann transfer orbit that would take the spacecraft from Orbit O to a coplanar circular orbit of radius $5.00\,\text{D}$ As such we can just use the formula for Impulse 1, but substitute in $r_1 = 1.00 \text{D}$ and $r_2 = 5.00\,\text{D}$, and ignore Impulse 2 entirely.

$$\Delta v_{E} = 0.29\,\text{D/T}$$

Results

In this counterexample case, the delta-V required to go from Orbit O to orbit C is $0.10\,\text{D/T}$ more than for going from Orbit O to Orbit E. As a result, all other things about the rocket being the same on the two different transfers, in this case, the rocket would require more fuel to go to Orbit C.

As a result we can conclude that for two different orbits of the same arbitrarily-chosen semi-major axis, there is no expectation that the fuel costs to reach both be the same.

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    $\begingroup$ So the reason why a circular orbit requires more delta-v is because you have to circularize it in the first place (once you reach apogee)? Isn't it possible to reach a very low (approximately-)circular orbit right away, e.g. at 100 mi (160 km) above the Earth's surface, without the need to adjust a too elliptical orbit? Your parking orbit O could be such a (near-)circular orbit that might be reached right away, couldn't it? $\endgroup$
    – Johannes
    Jul 23, 2023 at 13:00
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    $\begingroup$ Basically...it's complicated. To demonstrate that there's no requirement that the fuel expenditure be equal, I chose an easy-to-calculate situation that showed that the fuel expenditures would be different. With the infinite number of coplanar elliptical orbits with the same semi-major axis as the circular, there are absolutely situations where getting into the chosen elliptical orbit takes more delta-V. $\endgroup$
    – notovny
    Jul 23, 2023 at 13:28
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    $\begingroup$ I love "Basically...it's complicated"! I must now go use that phrase myself. $\endgroup$
    – Ryan C
    Jul 24, 2023 at 4:09

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