hi
how can i solved this problem ...
by using integrating of area proof
the ratio between Aphelion velocity and Perihelion velocity equal the ratio between both of them radius
if The orbit is an ellipse???
I need your help as soon as you can
hi
how can i solved this problem ...
by using integrating of area proof
the ratio between Aphelion velocity and Perihelion velocity equal the ratio between both of them radius
if The orbit is an ellipse???
I need your help as soon as you can
If you mean:Originally Posted by sweet
Using Kepler's second law - that the area swept out in unit time by a planet
is a constant - then yes.
At Aphelion and Perihelion the velocity of the planet is at right angles to the
vector from the Sun to the Planet, so the area swept out in unit time in
each case is:
$\displaystyle
\frac{d}{dt}A_{Ap}=v_{Ap}\ r_{Ap}/2
$
and:
$\displaystyle
\frac{d}{dt}A_{Peri}=v_{Peri}\ r_{Peri}/2
$
at Aphelion and Perihelion respectivly. Kepler's second law tells us these are
equal so:
$\displaystyle
\frac{d}{dt}A_{Ap}=\frac{d}{dt}A_{Peri}
$,
so:
$\displaystyle
v_{Ap}\ r_{Ap}/2=v_{Peri}\ r_{Peri}/2
$
Hence:
$\displaystyle
\frac{v_{Ap}}{v_{Peri}}=\frac{r_{Peri}}{v_{Ap}}
$
RonL
If I understand correctly you need to show that:Originally Posted by sweet
$\displaystyle \frac{v_{Aph}}{v_{Per}}=\frac{r_{Aph}}{r_{Per}}$
Unfortunately it isn't true. So I hope I misunderstood your question!
This is an interesting problem, since my book does not give a formula for the speed during an orbit.
For starters, I'll assume you are referring to an inverse-square force law (ie. the "Kepler Problem". Note that it is possible to produce elliptical orbits from other force laws, so this is necessary information.
The position in terms of the eccentricity is:
$\displaystyle r = \frac{a(1-e^2)}{1+e \, cos \theta}$
where a is the semi-major axis $\displaystyle a = \frac{r_{Aph}+r_{Per}}{2}=-\frac{GM}{2E}$ where E is the energy of the body in orbit, M is the central mass, and G is the Universal Gravitation constant.
For perihelion we set $\displaystyle \theta = 0 \,rad $ and for aphelion we set $\displaystyle \theta = \pi \, rad$. This gives a ratio of the radii as:
$\displaystyle \frac{r_{Aph}}{r_{Per}} = \frac{\frac{a(1-e^2)}{1-e}}{\frac{a(1-e^2)}{1+e}}$ = $\displaystyle \frac{1+e}{1-e}$
Now we need the apsidal speeds. As with all velocities, the direction is tangent to the path traced by the object over time. Hence we are looking for $\displaystyle r \frac{d \theta}{dt}$.
We don't know $\displaystyle d \theta /dt$ and we aren't going to get it. However, we do know a relation involving both it and the (constant) angular momentum:
$\displaystyle \frac{d \theta}{dt} = \frac{L}{mr^2}$ where m is the reduced mass of the system, and L is the angular momentum.
So at aphelion $\displaystyle r \frac{d \theta}{dt} = \frac{L(1-e)}{ma(1-e^2)}$.
and at perihelion $\displaystyle r \frac{d \theta}{dt} = \frac{L(1+e)}{ma(1-e^2)}$.
Thus:
$\displaystyle \frac{v_{Aph}}{v_{Per}}$ = $\displaystyle \frac{\frac{L(1-e)}{ma(1-e^2)}}{\frac{L(1+e)}{ma(1-e^2)}}$ = $\displaystyle \frac{1-e}{1+e}$
which is not the same as the expression we want. It is, however, the reciprocal of the quotient of the radii, so I'm hoping I understood your question incorrectly.
-Dan
$\displaystyle \frac{d}{dt}A_{Ap}$ is the rate that area is swept out by a line segment conecting the Sun to PlanetOriginally Posted by sweet
when the Planet is at Aphelion. Not a strictly correct notation, perhaps $\displaystyle \left \frac{dA}{dt} \right| _{Ap}$ would be better.
RonL