0
$\begingroup$

I am trying to model a rocket orbiting a planet. I want to simplify the problem so that the rocket is travelling along an infinitely long flat surface with a $g$ along the $y$-axis.

This is what I have done. Suppose the rocket is orbiting the planet. The relationship between $r$ (distance to the center of the planet), $v$ (the speed perpendicular to the line from the rocket to the center of the planet) and $g$ (gravitational acceleration) is $$g = \frac{v^2}{r}$$ If the rocket is powered on for a short while, the speed will decrease, and we "need" a smaller g for it to follow a circular orbit path. So I have set up this relationship. Here, $g_S$ is the gravitational acceleration in the simplified model:

$$ g - g_S = \frac{v^2}{r} $$ $$ g_S = -\frac{v^2}{r} + g $$ $$ g_S = -\frac{v^2}{r} + \frac{\gamma M}{r^2} $$ We multiply with the mass to get the gravitational force. $M$ is the mass of the planet. $$ G_S = -\frac{mv^2}{r} + \frac{\gamma Mm}{r^2} $$

enter image description here

Is my reasoning correct? It all depends on whether or not the function $g = v^2 / r$ applies in this situation.

$\endgroup$
  • $\begingroup$ I do not understand why you did not used $g-g_F=\frac{v^2}{r}-\frac{v_F^2}{r_F}$ $\endgroup$ – Wolphram jonny Oct 22 '14 at 17:53
  • $\begingroup$ @julianfernandez Where did you get this formula from? $\endgroup$ – Friend of Kim Oct 22 '14 at 19:24
  • $\begingroup$ from your own definition. if $g=\frac{v^2}{r}$ then $g_F$ should be $\frac{v_F^2}{r_F}$ $\endgroup$ – Wolphram jonny Oct 22 '14 at 19:55
  • $\begingroup$ @julianfernandez The formula $g=\frac{v^2}{r}$ doesn't apply that way (if I'm not mistaken). The formula states what the gravitational acceleration must be to achieve a circular orbit. I.e. $v = \sqrt{gr} = \sqrt{\frac{\gamma M}{r}}$ is the velocity an object needs to have to orbit a planet with mass $M$ $r$ meters above the center of the planet. $\endgroup$ – Friend of Kim Oct 22 '14 at 23:13
  • $\begingroup$ Hi Friend of Kim. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Oct 22 '14 at 23:38
1
$\begingroup$

You are basically asking how to describe the equations of motion when viewed from an rotating frame of reference (the planets surface). This can be done by including the Coriolis effect.

For example if you simplify the problem to just the equator of the planet the equations of motion would then become:

$$ \ddot{x}=\frac{F_x}{m}-\frac{v_xv_y}{r}, $$ $$ \ddot{y}=\frac{F_y}{m}+\frac{v_x^2}{r}-\frac{GM}{r^2}, $$ where $\ddot{x}$ is the acceleration on the horizontal axis, $\ddot{y}$ is the acceleration on the vertical axis, $F_i$ the component of the applied force (drag or thrust ect.) in the $i$-direction, $m$ is the mass of the rocket, $v_i$ the component of the velocity in the $i$-direction, $G$ the gravitational constant, $M$ the mass of the planet and $r$ the distance from the center of mass of the planet to the rocket. However do keep in mind that these velocities are relative to the orbital velocities. So if you want to calculate it relative to the surface velocity you would have to add $r\omega$ to $v_x$, where $\omega$ is the angular velocity of the planet (usually in radians per second).

$\endgroup$
  • $\begingroup$ So the acceleration in the $y$-direction is correct. What you are saying in addition to what I've suggested is that $\ddot{x}$ is also affected by the velocity? How did you get this equation? $\endgroup$ – Friend of Kim Oct 26 '14 at 14:47
  • $\begingroup$ @FriendofKim yes, your vertical component of acceleration is correct (however I defined up as the positive direction, while you use downwards). Did you have a look at the Coriolis effect, since this can be derived from that, by assuming that you have no velocity component in the north/south direction. $\endgroup$ – fibonatic Oct 26 '14 at 21:38
  • $\begingroup$ @FriendofKim an easy way to intuitively understand the term of in $\ddot{x}$ is that it helps to conserve the angular momentum (if the other applied forces would be zero). $\endgroup$ – fibonatic Oct 29 '14 at 3:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.