# can any body help me

• May 22nd 2006, 11:24 AM
sweet
can any body help me
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
• May 22nd 2006, 12:47 PM
CaptainBlack
Quote:

Originally Posted by sweet
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:

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:

$
\frac{d}{dt}A_{Ap}=v_{Ap}\ r_{Ap}/2
$

and:

$
\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:

$
\frac{d}{dt}A_{Ap}=\frac{d}{dt}A_{Peri}
$
,

so:

$
v_{Ap}\ r_{Ap}/2=v_{Peri}\ r_{Peri}/2
$

Hence:

$
\frac{v_{Ap}}{v_{Peri}}=\frac{r_{Peri}}{v_{Ap}}
$

RonL
• May 22nd 2006, 01:22 PM
topsquark
Quote:

Originally Posted by sweet
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 I understand correctly you need to show that:
$\frac{v_{Aph}}{v_{Per}}=\frac{r_{Aph}}{r_{Per}}$

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 &quot;Kepler Problem&quot;). 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:
$r = \frac{a(1-e^2)}{1+e \, cos \theta}$
where a is the semi-major axis $a = \frac{r_{Aph}+r_{Per}}{2}=-\frac{GM}{2E}$ 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 $\theta = 0 \,rad$ and for aphelion we set $\theta = \pi \, rad$. This gives a ratio of the radii as:

$\frac{r_{Aph}}{r_{Per}} = \frac{\frac{a(1-e^2)}{1-e}}{\frac{a(1-e^2)}{1+e}}$ = $\frac{1+e}{1-e}$

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 $r \frac{d \theta}{dt}$.

We don't know $d \theta /dt$ and we aren't going to get it. However, we do know a relation involving both it and the (constant) angular momentum:
$\frac{d \theta}{dt} = \frac{L}{mr^2}$ where m is the reduced mass of the system, and L is the angular momentum.

So at aphelion $r \frac{d \theta}{dt} = \frac{L(1-e)}{ma(1-e^2)}$.

and at perihelion $r \frac{d \theta}{dt} = \frac{L(1+e)}{ma(1-e^2)}$.

Thus:

$\frac{v_{Aph}}{v_{Per}}$ = $\frac{\frac{L(1-e)}{ma(1-e^2)}}{\frac{L(1+e)}{ma(1-e^2)}}$ = $\frac{1-e}{1+e}$

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
• May 22nd 2006, 01:43 PM
sweet
CaptainBlack

:D thank u soooo much

but what u mean about A_{Ap}

if u have some paper about Kepler's problem and two body problem

plese give it to me i will grateful for u
• May 22nd 2006, 01:47 PM
sweet
thank u topsquark
u r method very long but i will think about it

:D
• May 22nd 2006, 01:55 PM
CaptainBlack
Quote:

Originally Posted by sweet
CaptainBlack

:D thank u soooo much

but what u mean about A_{Ap}

if u have some paper about Kepler's problem and two body problem

plese give it to me i will grateful for u

$\frac{d}{dt}A_{Ap}$ is the rate that area is swept out by a line segment conecting the Sun to Planet
when the Planet is at Aphelion. Not a strictly correct notation, perhaps $\left \frac{dA}{dt} \right| _{Ap}$ would be better.

RonL
• May 22nd 2006, 02:17 PM
sweet
ok ...but i know

$
\frac{d}{dt}A=r^2 \frac{d}{dt}[thet] /2

$

and u writ it in the form
$
\frac{d}{dt}A_{Ap}=v_{Ap}\ r_{Ap}/2
$

??????
• May 22nd 2006, 02:26 PM
CaptainBlack
Quote:

Originally Posted by sweet
ok ...but i know

$
\frac{d}{dt}A=r^2 \frac{d}{dt}[thet] /2

$

and u writ it in the form
$
\frac{d}{dt}A_{Ap}=v_{Ap}\ r_{Ap}/2
$

??????

At aphelion:

$\frac{d}{dt}\theta=v_{Ap}/r_{Ap}$

because the Planet is moving at rightangles to the segment connecting the
Sun to the Planet.

RonL
• May 23rd 2006, 12:05 AM
sweet
i get it

CaptainBlack u r the best :cool: