I am asking this question in relation to an example shown in the 'Orbital Mechanics for Engineering Students' book by Howard Curtis (third edition) pp659 example 12.1.

The example is demonstrating how to calculate the orbit decay given the Keplerian elements and ballistic coefficient. The process goes something like this:

Convert Keplerians Elements ==> [R, V] State Vector in perifocal frame ==> [R, V] state vector in Geocentric equatorial frame ==> numerical solution of equation of motion:

$$\ddot{\mathbf{r}} = -\mu \frac{\mathbf{r}}{r^3} + p$$

where p is the perturbing drag acceleration:

$$\mathbf{p} = -\frac{1}{2} \rho \upsilon^2 \left(\frac{C_D A}{m} \right) \mathbf{v_{rel}}$$

After numerical solution of the ODE, a plot of altitude vs. time is generated.

My question is: why is it necessary to convert the Keplerian elements into a state vector in the geocentric equitorial frame. I get that a state vector representation is needed to work with the equations of motions, but why can't you use just solve the equations of motions with the state vectors in perifocal frame? Is there a particular reason/advantage of doing this?

  • $\begingroup$ It looks like this is an exercise meant to be educational. It doesn't really say it is necessary, does it? $\endgroup$
    – uhoh
    Mar 17, 2018 at 13:00
  • $\begingroup$ Also, I've changed your equations to MathJax which is the supported way to render equations here (rather than the external site you linked to). Are you sure the second equation is correct? Shouldn't it be just $\upsilon \mathbf{v}$ instead of $\upsilon^2 \mathbf{v}$? Check your units. $\endgroup$
    – uhoh
    Mar 17, 2018 at 13:15

1 Answer 1


I'm not a specialist, but here are my guesses: the geocentric equatorial frame makes it easier to express $\mathbf{v}_{rel}$, the spacecraft's velocity relative to the atmosphere, in terms of $\mathbf{r}$ and $\mathbf{v}$: $$\mathbf{v}_{rel} = \mathbf{v} - \boldsymbol{\omega}_E\times \mathbf{r}, $$ where $\boldsymbol{\omega}_E$ is the angular velocity of Earth's rotation, which points in the $z$-direction.

Well, if by perifocal frame you mean the one for the initial orbit, with the directions of the axes staying the same (I'm not sure whether that's how this term is used), then the formula for $\mathbf{v}_{rel}$ is the same. The only difference is that $\boldsymbol{\omega}_E$ doesn't point in the direction of an axis anymore, you need to calculate its direction first. But sure, you can solve the equations in this frame as well. Note, however, that since $\mathbf{v}_{rel}$ and the drag don't have to lie in the orbit's plane, the plane may change over time, which decreases the usefulness of the perifocal frame (it won't be perifocal for later orbits).

And if you want a frame whose axes change directions to keep track of the orbit's orientation, then the equations become much more complicated.


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