Conversation started Oct 10, 2019 at 5:22.
Oct 10, 2019 05:22
@Nobodyrecognizeable hi
@JohnRennie ^^
Suppose the apogee of the satellite is at a distance $d$. You know that energy is conserved,so the total energy (PE + KE) of the satellite is constant. You can calculate the initial KE and PE because you're given the initial distance and velocity. OK so far?
@JohnRennie OK.
And you can calculate the PE at the apogee because it is just $U = -GMm/d$. So if you can find the KE at the apogee you can calculate $d$. Yes?
@JohnRennie yes.
Oct 10, 2019 05:30
And the way you calculate $d$ is to use conservation of angular momentum. At launch the velocity is tangent to the radius vector so the angular momentum is just $L = v_i^2/R$ where $v_i$ is the initial velocity.
At apogee the velocity is again normal to the radius vector $d$ so again $L=v_a^2/d$ where $v_a$ is the apogee velocity.
@JohnRennie OK.
So conservation of energy gives you one equation in $d$ and $v_a$, and conservation of angular momentum gives you a second equation in $d$and $v_a$. So you have two equations in two variables. Solve these to get $d$ and $v_a$.
What you are doing here is an application of the vis-viva equation i.e. using conservation of energy and angular momentum:
In astrodynamics, the vis-viva equation, also referred to as orbital-energy-invariance law, is one of the equations that model the motion of orbiting bodies. It is the direct result of the principle of conservation of mechanical energy which applies when the only force acting on an object is its own weight. Vis viva (Latin for "living force") is a term from the history of mechanics, and it survives in this sole context. It represents the principle that the difference between the total work of the accelerating forces of a system and that of the retarding forces is equal to one half the vis viva...
@JohnRennie I get b.
@JohnRennie got it now sir :-) Thank you very much for helping sir
@Nobodyrecognizeable does it match the answer key?
Oct 10, 2019 05:40
@JohnRennie yes.
@Nobodyrecognizeable BOOM! :-)
It was easier than you thought :-)
@JohnRennie from vis viva equation we are finding a right?
@JohnRennie certainly.
The vis-viva equation wouldn't answer your question. I mention it because it is derived using exactly the method you just used to answer the question.
@JohnRennie aren't we finding a?
No, because $a$ is the semimajor axis not the apogee distance.
If $d_a$ and $d_p$ are the apogee and perigee distances then $2a = d_a+ d_p$
Oct 10, 2019 05:46
@JohnRennie I get a ad R and d_a =2r so the problem is unrealistic.
You can calculate the perigee distance in exactly the same way you just calculated the apogee distance, though in this case the perigee distance would turn out to be below the surface of the Earth.
Oh,no wait, the perigee distance would just be $R$.
So you'd get $a = 1.5R$
No, sorry, wrong again.
@JohnRennie don't we take the sun stationary? So a=2r
The answer is the distance above the earth not the distance from the centre of the Earth. The apogee distance from the centre of the earth is $d_a = 3R$.
@JohnRennie won't that mean a circular orbit?
And the perigee distance is just $R$ so the distance between the apogee and perigee points is $d_a + d_p = 4R$. Yes?
Oct 10, 2019 05:51
@JohnRennie so indeed from surface of earth its 2r.
Ya I get $d_a=3r $ and $d_p=r$ when you measure from the centre of earth. Have to subtract r if you wanna get from surface.
@JohnRennie ^^
In the vis-viva equation the distance $r$ is the distance from the centre of the Earth.
You almost never use the distance from the surface in orbital mechanics.
 
Conversation ended Oct 10, 2019 at 5:55.