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Calculate xyz velocity vectors for circular orbit
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I'm trying to calculate the xyz velocity vectors for a circular orbit given a set of position vectors, but I can't seem to get it right.
The formula for obtaining the velocity magnitude is the square root of gravitational constant multiplied by the mass of the primary divided by the distance from the primary to the object orbiting it. I'm using solar masses for mass, astronomical units for distance and years for time. To simplify things, the primary is our sun, which means that G * M is equal to 39.5. The position of the Sun is 0, 0, 0 and the vectors you see for Earth are in au and were obtained from the JPL Horizons tool.
Here's the code:
function getDistanceParams(p1, p2)
const dx = p2.x - p1.x;
const dy = p2.y - p1.y;
const dz = p2.z - p1.z;
return dx, dy, dz, dSquared: dx * dx + dy * dy + dz * dz ;
function getCircularOrbit()
const dParams = getDistanceParams(
x: 0, y: 0, z: 0 ,
x: 0.9197324105567349,
y: -0.4147318273536994,
z: -1.750037390352759e-5
);
const vMag = Math.sqrt(39.5 / Math.sqrt(dParams.dSquared));
return vx: dParams.dx * vMag, vy: dParams.dy * vMag, vz: dParams.dz * vMag;
getCircularOrbit();
/*
The above outputs this:
vx: 5.754832301294686,
vy: -2.59500707927136,
vz: -0.00010950110691846996
The actual velocity vectors:
vx: 2.4645428337894026,
vy: 5.7097644117945805,
vz: -3.3177815766459033e-4,
*/
Where am I going wrong???
vector
asked 3 hours ago
Happy Koala
1135
add a comment |Â
up vote
1
down vote
favorite
I'm trying to calculate the xyz velocity vectors for a circular orbit given a set of position vectors, but I can't seem to get it right.
The formula for obtaining the velocity magnitude is the square root of gravitational constant multiplied by the mass of the primary divided by the distance from the primary to the object orbiting it. I'm using solar masses for mass, astronomical units for distance and years for time. To simplify things, the primary is our sun, which means that G * M is equal to 39.5. The position of the Sun is 0, 0, 0 and the vectors you see for Earth are in au and were obtained from the JPL Horizons tool.
Here's the code:
function getDistanceParams(p1, p2)
const dx = p2.x - p1.x;
const dy = p2.y - p1.y;
const dz = p2.z - p1.z;
return dx, dy, dz, dSquared: dx * dx + dy * dy + dz * dz ;
function getCircularOrbit()
const dParams = getDistanceParams(
x: 0, y: 0, z: 0 ,
x: 0.9197324105567349,
y: -0.4147318273536994,
z: -1.750037390352759e-5
);
const vMag = Math.sqrt(39.5 / Math.sqrt(dParams.dSquared));
return vx: dParams.dx * vMag, vy: dParams.dy * vMag, vz: dParams.dz * vMag;
getCircularOrbit();
/*
The above outputs this:
vx: 5.754832301294686,
vy: -2.59500707927136,
vz: -0.00010950110691846996
The actual velocity vectors:
vx: 2.4645428337894026,
vy: 5.7097644117945805,
vz: -3.3177815766459033e-4,
*/
Where am I going wrong???
vector
asked 3 hours ago
Happy Koala
1135
Can you write the math equations here? After we (or math stackexchange) check that, go to stackoverflow to check if your programming is same as the math equation.
– R zu
3 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to calculate the xyz velocity vectors for a circular orbit given a set of position vectors, but I can't seem to get it right.
The formula for obtaining the velocity magnitude is the square root of gravitational constant multiplied by the mass of the primary divided by the distance from the primary to the object orbiting it. I'm using solar masses for mass, astronomical units for distance and years for time. To simplify things, the primary is our sun, which means that G * M is equal to 39.5. The position of the Sun is 0, 0, 0 and the vectors you see for Earth are in au and were obtained from the JPL Horizons tool.
Here's the code:
function getDistanceParams(p1, p2)
const dx = p2.x - p1.x;
const dy = p2.y - p1.y;
const dz = p2.z - p1.z;
return dx, dy, dz, dSquared: dx * dx + dy * dy + dz * dz ;
function getCircularOrbit()
const dParams = getDistanceParams(
x: 0, y: 0, z: 0 ,
x: 0.9197324105567349,
y: -0.4147318273536994,
z: -1.750037390352759e-5
);
const vMag = Math.sqrt(39.5 / Math.sqrt(dParams.dSquared));
return vx: dParams.dx * vMag, vy: dParams.dy * vMag, vz: dParams.dz * vMag;
getCircularOrbit();
/*
The above outputs this:
vx: 5.754832301294686,
vy: -2.59500707927136,
vz: -0.00010950110691846996
The actual velocity vectors:
vx: 2.4645428337894026,
vy: 5.7097644117945805,
vz: -3.3177815766459033e-4,
*/
Where am I going wrong???
vector
asked 3 hours ago
Happy Koala
1135
I'm trying to calculate the xyz velocity vectors for a circular orbit given a set of position vectors, but I can't seem to get it right.
The formula for obtaining the velocity magnitude is the square root of gravitational constant multiplied by the mass of the primary divided by the distance from the primary to the object orbiting it. I'm using solar masses for mass, astronomical units for distance and years for time. To simplify things, the primary is our sun, which means that G * M is equal to 39.5. The position of the Sun is 0, 0, 0 and the vectors you see for Earth are in au and were obtained from the JPL Horizons tool.
Here's the code:
function getDistanceParams(p1, p2)
const dx = p2.x - p1.x;
const dy = p2.y - p1.y;
const dz = p2.z - p1.z;
return dx, dy, dz, dSquared: dx * dx + dy * dy + dz * dz ;
function getCircularOrbit()
const dParams = getDistanceParams(
x: 0, y: 0, z: 0 ,
x: 0.9197324105567349,
y: -0.4147318273536994,
z: -1.750037390352759e-5
);
const vMag = Math.sqrt(39.5 / Math.sqrt(dParams.dSquared));
return vx: dParams.dx * vMag, vy: dParams.dy * vMag, vz: dParams.dz * vMag;
getCircularOrbit();
/*
The above outputs this:
vx: 5.754832301294686,
vy: -2.59500707927136,
vz: -0.00010950110691846996
The actual velocity vectors:
vx: 2.4645428337894026,
vy: 5.7097644117945805,
vz: -3.3177815766459033e-4,
*/
Where am I going wrong???
vector
vector
asked 3 hours ago
Happy Koala
1135
asked 3 hours ago
Happy Koala
1135
edited 3 hours ago
asked 3 hours ago
Happy Koala
1135
asked 3 hours ago
Happy Koala
1135
asked 3 hours ago
Happy Koala
1135
1135
Can you write the math equations here? After we (or math stackexchange) check that, go to stackoverflow to check if your programming is same as the math equation.
– R zu
3 hours ago
add a comment |Â
Can you write the math equations here? After we (or math stackexchange) check that, go to stackoverflow to check if your programming is same as the math equation.
– R zu
3 hours ago
Can you write the math equations here? After we (or math stackexchange) check that, go to stackoverflow to check if your programming is same as the math equation.
– R zu
3 hours ago
Can you write the math equations here? After we (or math stackexchange) check that, go to stackoverflow to check if your programming is same as the math equation.
– R zu
3 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
In your code, you seem to be calculating the magnitude $|vecv|$ of the velocity (the speed) from your gravitational formula and multiplying it by the components of the position vector $vecr$. There are two errors here.
- You need to multiply the speed by a unit vector in the direction of the velocity, in order to get the velocity vector $vecv$. In general the
position vector is not a unit vector (actually in this case, it is pretty close, because you chose astronomical units, but that is just luck). - The position vector $vecr$ is pointing in a radial direction, whereas the velocity vector $vecv$ (for circular motion)
lies in a tangential direction.
You are not actually inputting to your program enough information to work out the direction of the velocity vector. You need to know the plane in which the rotation is taking place, and the sense of the rotation (clockwise or anticlockwise). The key equation is
$$
vecv=vecomegatimesvecr=omegahatntimesvecr
$$
where $vecomega=omegahatn$ is the angular velocity vector
and $times$ is the vector cross product.
Assuming that you know the unit vector $hatn$
pointing in the direction of
$vecomega$,
you can deduce the direction of $vecv$ from the above formula.
Alternatively, you can calculate the magnitude $omega$ from the speed
and the radius of the orbit, and use it together with $hatn$
directly in the above formula.
Of course, if you are prepared to make the approximation that the rotation
is entirely in the $xy$ plane (which is not exactly correct, from your numbers, but quite close) then you know $hatn$ and can easily work out the direction of $vecv$ from $vecr$.
answered 2 hours ago
LonelyProf
42116
Thank you! Silly of me not to think of the direction in which the body rotates.
– Happy Koala
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
In your code, you seem to be calculating the magnitude $|vecv|$ of the velocity (the speed) from your gravitational formula and multiplying it by the components of the position vector $vecr$. There are two errors here.
- You need to multiply the speed by a unit vector in the direction of the velocity, in order to get the velocity vector $vecv$. In general the
position vector is not a unit vector (actually in this case, it is pretty close, because you chose astronomical units, but that is just luck). - The position vector $vecr$ is pointing in a radial direction, whereas the velocity vector $vecv$ (for circular motion)
lies in a tangential direction.
You are not actually inputting to your program enough information to work out the direction of the velocity vector. You need to know the plane in which the rotation is taking place, and the sense of the rotation (clockwise or anticlockwise). The key equation is
$$
vecv=vecomegatimesvecr=omegahatntimesvecr
$$
where $vecomega=omegahatn$ is the angular velocity vector
and $times$ is the vector cross product.
Assuming that you know the unit vector $hatn$
pointing in the direction of
$vecomega$,
you can deduce the direction of $vecv$ from the above formula.
Alternatively, you can calculate the magnitude $omega$ from the speed
and the radius of the orbit, and use it together with $hatn$
directly in the above formula.
Of course, if you are prepared to make the approximation that the rotation
is entirely in the $xy$ plane (which is not exactly correct, from your numbers, but quite close) then you know $hatn$ and can easily work out the direction of $vecv$ from $vecr$.
answered 2 hours ago
LonelyProf
42116
Thank you! Silly of me not to think of the direction in which the body rotates.
– Happy Koala
2 hours ago
add a comment |Â
up vote
2
down vote
accepted
In your code, you seem to be calculating the magnitude $|vecv|$ of the velocity (the speed) from your gravitational formula and multiplying it by the components of the position vector $vecr$. There are two errors here.
- You need to multiply the speed by a unit vector in the direction of the velocity, in order to get the velocity vector $vecv$. In general the
position vector is not a unit vector (actually in this case, it is pretty close, because you chose astronomical units, but that is just luck). - The position vector $vecr$ is pointing in a radial direction, whereas the velocity vector $vecv$ (for circular motion)
lies in a tangential direction.
You are not actually inputting to your program enough information to work out the direction of the velocity vector. You need to know the plane in which the rotation is taking place, and the sense of the rotation (clockwise or anticlockwise). The key equation is
$$
vecv=vecomegatimesvecr=omegahatntimesvecr
$$
where $vecomega=omegahatn$ is the angular velocity vector
and $times$ is the vector cross product.
Assuming that you know the unit vector $hatn$
pointing in the direction of
$vecomega$,
you can deduce the direction of $vecv$ from the above formula.
Alternatively, you can calculate the magnitude $omega$ from the speed
and the radius of the orbit, and use it together with $hatn$
directly in the above formula.
Of course, if you are prepared to make the approximation that the rotation
is entirely in the $xy$ plane (which is not exactly correct, from your numbers, but quite close) then you know $hatn$ and can easily work out the direction of $vecv$ from $vecr$.
answered 2 hours ago
LonelyProf
42116
Thank you! Silly of me not to think of the direction in which the body rotates.
– Happy Koala
2 hours ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
In your code, you seem to be calculating the magnitude $|vecv|$ of the velocity (the speed) from your gravitational formula and multiplying it by the components of the position vector $vecr$. There are two errors here.
- You need to multiply the speed by a unit vector in the direction of the velocity, in order to get the velocity vector $vecv$. In general the
position vector is not a unit vector (actually in this case, it is pretty close, because you chose astronomical units, but that is just luck). - The position vector $vecr$ is pointing in a radial direction, whereas the velocity vector $vecv$ (for circular motion)
lies in a tangential direction.
You are not actually inputting to your program enough information to work out the direction of the velocity vector. You need to know the plane in which the rotation is taking place, and the sense of the rotation (clockwise or anticlockwise). The key equation is
$$
vecv=vecomegatimesvecr=omegahatntimesvecr
$$
where $vecomega=omegahatn$ is the angular velocity vector
and $times$ is the vector cross product.
Assuming that you know the unit vector $hatn$
pointing in the direction of
$vecomega$,
you can deduce the direction of $vecv$ from the above formula.
Alternatively, you can calculate the magnitude $omega$ from the speed
and the radius of the orbit, and use it together with $hatn$
directly in the above formula.
Of course, if you are prepared to make the approximation that the rotation
is entirely in the $xy$ plane (which is not exactly correct, from your numbers, but quite close) then you know $hatn$ and can easily work out the direction of $vecv$ from $vecr$.
answered 2 hours ago
LonelyProf
42116
In your code, you seem to be calculating the magnitude $|vecv|$ of the velocity (the speed) from your gravitational formula and multiplying it by the components of the position vector $vecr$. There are two errors here.
- You need to multiply the speed by a unit vector in the direction of the velocity, in order to get the velocity vector $vecv$. In general the
position vector is not a unit vector (actually in this case, it is pretty close, because you chose astronomical units, but that is just luck). - The position vector $vecr$ is pointing in a radial direction, whereas the velocity vector $vecv$ (for circular motion)
lies in a tangential direction.
You are not actually inputting to your program enough information to work out the direction of the velocity vector. You need to know the plane in which the rotation is taking place, and the sense of the rotation (clockwise or anticlockwise). The key equation is
$$
vecv=vecomegatimesvecr=omegahatntimesvecr
$$
where $vecomega=omegahatn$ is the angular velocity vector
and $times$ is the vector cross product.
Assuming that you know the unit vector $hatn$
pointing in the direction of
$vecomega$,
you can deduce the direction of $vecv$ from the above formula.
Alternatively, you can calculate the magnitude $omega$ from the speed
and the radius of the orbit, and use it together with $hatn$
directly in the above formula.
Of course, if you are prepared to make the approximation that the rotation
is entirely in the $xy$ plane (which is not exactly correct, from your numbers, but quite close) then you know $hatn$ and can easily work out the direction of $vecv$ from $vecr$.
answered 2 hours ago
LonelyProf
42116
answered 2 hours ago
LonelyProf
42116
answered 2 hours ago
LonelyProf
42116
answered 2 hours ago
LonelyProf
42116
42116
Thank you! Silly of me not to think of the direction in which the body rotates.
– Happy Koala
2 hours ago
add a comment |Â
Thank you! Silly of me not to think of the direction in which the body rotates.
– Happy Koala
2 hours ago
Thank you! Silly of me not to think of the direction in which the body rotates.
– Happy Koala
2 hours ago
Thank you! Silly of me not to think of the direction in which the body rotates.
– Happy Koala
2 hours ago
add a comment |Â
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Can you write the math equations here? After we (or math stackexchange) check that, go to stackoverflow to check if your programming is same as the math equation.
– R zu
3 hours ago