- #1

- 126

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter DiamondGeezer
- Start date

- #1

- 126

- 0

- #2

- 15,415

- 687

- #3

- 126

- 0

Well it might be easy but I don't know.

There is one way to get the tangential velocity directly by using the orbital period T, so that

[tex]v_{tangent}=\frac{2 \pi r}{T}[/tex]

but I'd like to work it out from the velocity vectors given.

As far as I can tell, the JPL Horizons program produces coordinate axes in fixed directions rather than rotating with the orbiting body. It does produce a range rate (RR) parameter but I'd still like to know how to utilize the data WITHOUT referencing the radius from the origin, since that is what I'm trying to measure as accurately as I can.

So if someone knows how to produce tangential velocity from the three velocity vectors + RR (or some other equivalent method) then I'm all ears.

- #4

- 15,415

- 687

[tex]{\boldsymbol b}_a =

\frac{({\boldsymbol b}\cdot{\boldsymbol a})\, {\boldsymbol a}}

{||{\boldsymbol a}||^2}[/tex]

The component of

[tex]{\boldsymbol b}_{\perp} =

{\boldsymbol b} -

\frac{({\boldsymbol b}\cdot{\boldsymbol a})\, {\boldsymbol a}}

{||{\boldsymbol a}||^2}[/tex]

Alternatively you can use the vector triple product:

[tex]{\boldsymbol b}_{\perp} =

\frac{{\boldsymbol a}\times({\boldsymbol b}\times{\boldsymbol a})}

{||{\boldsymbol a}||^2}[/tex]

- #5

- 126

- 0

From the Horizons vector output, you have x-y-z position of Sun wrt object, so

can get the radial vector. Normalize it to get the unit vector basis u1.

The velocity vector is in the plane of motion. Normalize it to get v1, then

compute the cross-product v1 x u1 to get 2nd basis vector out of the plane, u2.

Then compute the cross product of the radial (in-plane) and out-of-plane

basis vectors to get a third vector in the plane: u1 x u2 = u3

u3 defines the tangential direction. Resolve the original velocity

vector along it (dot product) to get the tangential component of velocity.

Well that's fine but I need to work out from an example:

Here is the data from one particular time for a spacecraft relative to the Sun:

2453736.875000000, A.D. 2006-Jan-01 09:00:00.0000, -2.156346497896881E+07, 1.320794658183279E+08, -4.625147324320897E+05, -2.902293022684082E+01, -9.759936201814883E+00, -6.652160516632621E-02, 4.464052560742120E+02, 1.338289289826075E+08, -4.955740573003551E+00

where the data is

JDCT Epoch Julian Date, Coordinate Time

X x-component of position vector (km)

Y y-component of position vector (km)

Z z-component of position vector (km)

VX x-component of velocity vector (km/sec)

VY y-component of velocity vector (km/sec)

VZ z-component of velocity vector (km/sec)

LT One-way down-leg Newtonian light-time (sec)

RG Range; distance from coordinate center (km)

RR Range-rate; radial velocity wrt coord. center (km/sec)

in that order.

Q1: How do I calculate u1, u2, u3?

Q2: The radial velocity is RR, can we confirm that with a calculation?

- #6

- 126

- 0

[tex]p=(x,y,z) = (-2.156346497896881E+07, 1.320794658183279E+08, -4.625147324320897E+05)[/tex]

Therefore

[tex] |p|= \sqrt{x^2+y^2+z^2} [/tex]

therefore the unit radial vector is

[tex]u_1 = \frac{p}{|p|} [/tex]

To be continued...

- #7

- 126

- 0

Therefore the radial velocity is

[tex] v_{rad} = v \cdot u_1 [/tex]

[tex] v_{rad} = v \cdot u_1 [/tex]

- #8

- 126

- 0

therefore [tex]u_1= \frac{1}{1.33828928982607E+8} (-2.156346497896881E+07, 1.320794658183279E+08, -4.625147324320897E+05)[/tex]

- #9

- 126

- 0

And now to cross products...

- #10

- 126

- 0

[tex]v_1= \frac{v}{|v|}[/tex]

[tex]u_2 = v_1 \times u_1 [/tex]

[tex]u_3 = u_1 \times u_2[/tex]

Then [tex]v_{tangent}= v \cdot u_3[/tex] which I calculate to be 2.98180455442852 x 10^1 km/s

If somebody could check this, I would be very grateful....

Share: