Mixing
How to determine what direction an object will orbit another object.
$begingroup$
In this universe lets say everything exists in a 2d plane. There exists a 2d planet and hurdling towards it (at some velocity), is a 2d asteroid. Luckily, the asteroid isn't going to collide with the planet, however, the asteroid is going to get caught by the planets gravity and begin orbiting the planet.
Given 3 variables:
- Magnitude of the asteroid
- Position of the asteroid (in x, y
coordinates) - Position of the planet (in x, y, coordinates)
How can I determine if the asteroid will orbit clockwise or counter-clockwise?
Here are examples of the asteroid approaching the planet and the direction they would orbit.
physics
asked Jan 25 at 3:57
Gary HolidayGary Holiday
101
$endgroup$
add a comment |
$begingroup$
In this universe lets say everything exists in a 2d plane. There exists a 2d planet and hurdling towards it (at some velocity), is a 2d asteroid. Luckily, the asteroid isn't going to collide with the planet, however, the asteroid is going to get caught by the planets gravity and begin orbiting the planet.
Given 3 variables:
- Magnitude of the asteroid
- Position of the asteroid (in x, y
coordinates) - Position of the planet (in x, y, coordinates)
How can I determine if the asteroid will orbit clockwise or counter-clockwise?
Here are examples of the asteroid approaching the planet and the direction they would orbit.
physics
asked Jan 25 at 3:57
Gary HolidayGary Holiday
101
$endgroup$
1
$begingroup$
The vector product?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 4:00
$begingroup$
Can you elaborate more?
$endgroup$
– Gary Holiday
Jan 25 at 4:01
$begingroup$
What's point 1, magnitude of asteroid supposed to mean?
$endgroup$
– Ankit Kumar
Jan 25 at 4:21
add a comment |
$begingroup$
In this universe lets say everything exists in a 2d plane. There exists a 2d planet and hurdling towards it (at some velocity), is a 2d asteroid. Luckily, the asteroid isn't going to collide with the planet, however, the asteroid is going to get caught by the planets gravity and begin orbiting the planet.
Given 3 variables:
- Magnitude of the asteroid
- Position of the asteroid (in x, y
coordinates) - Position of the planet (in x, y, coordinates)
How can I determine if the asteroid will orbit clockwise or counter-clockwise?
Here are examples of the asteroid approaching the planet and the direction they would orbit.
physics
asked Jan 25 at 3:57
Gary HolidayGary Holiday
101
$endgroup$
In this universe lets say everything exists in a 2d plane. There exists a 2d planet and hurdling towards it (at some velocity), is a 2d asteroid. Luckily, the asteroid isn't going to collide with the planet, however, the asteroid is going to get caught by the planets gravity and begin orbiting the planet.
Given 3 variables:
- Magnitude of the asteroid
- Position of the asteroid (in x, y
coordinates) - Position of the planet (in x, y, coordinates)
How can I determine if the asteroid will orbit clockwise or counter-clockwise?
Here are examples of the asteroid approaching the planet and the direction they would orbit.
physics
physics
asked Jan 25 at 3:57
Gary HolidayGary Holiday
101
asked Jan 25 at 3:57
Gary HolidayGary Holiday
101
asked Jan 25 at 3:57
Gary HolidayGary Holiday
101
asked Jan 25 at 3:57
Gary HolidayGary Holiday
101
asked Jan 25 at 3:57
Gary HolidayGary Holiday
101
101
1
$begingroup$
The vector product?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 4:00
$begingroup$
Can you elaborate more?
$endgroup$
– Gary Holiday
Jan 25 at 4:01
$begingroup$
What's point 1, magnitude of asteroid supposed to mean?
$endgroup$
– Ankit Kumar
Jan 25 at 4:21
add a comment |
1
$begingroup$
The vector product?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 4:00
$begingroup$
Can you elaborate more?
$endgroup$
– Gary Holiday
Jan 25 at 4:01
$begingroup$
What's point 1, magnitude of asteroid supposed to mean?
$endgroup$
– Ankit Kumar
Jan 25 at 4:21
1
1
$begingroup$
The vector product?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 4:00
$begingroup$
The vector product?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 4:00
$begingroup$
Can you elaborate more?
$endgroup$
– Gary Holiday
Jan 25 at 4:01
$begingroup$
Can you elaborate more?
$endgroup$
– Gary Holiday
Jan 25 at 4:01
$begingroup$
What's point 1, magnitude of asteroid supposed to mean?
$endgroup$
– Ankit Kumar
Jan 25 at 4:21
$begingroup$
What's point 1, magnitude of asteroid supposed to mean?
$endgroup$
– Ankit Kumar
Jan 25 at 4:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You need the initial velocity of the asteroid (relative to a frame reference centered on the planet). If the asteroid were initially at rest in that frame, since gravity acts in the direction to the center of mass of the planet, the asteroid would follow a direct collision trajectory (a free fall on the planet).
If you have the velocity vector for the asteroid in the reference frame centered at the planet, then sign of the third component of the vector product of the position and velocity vectors (taken as vectors in $mathbb R^3$) predicts the orientation of the orbit. If you want to keep it two dimensional, think of the determinant of the matrix whose columns are the position and the velocity... again, the sign defines the orientation (which orientation is + and which is - depends on the detailed definition used for those vectors and the order in which they're taken for the calculations.)
answered Jan 25 at 4:01
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
You need the initial velocity of the asteroid (relative to a frame reference centered on the planet). If the asteroid were initially at rest in that frame, since gravity acts in the direction to the center of mass of the planet, the asteroid would follow a direct collision trajectory (a free fall on the planet).
If you have the velocity vector for the asteroid in the reference frame centered at the planet, then sign of the third component of the vector product of the position and velocity vectors (taken as vectors in $mathbb R^3$) predicts the orientation of the orbit. If you want to keep it two dimensional, think of the determinant of the matrix whose columns are the position and the velocity... again, the sign defines the orientation (which orientation is + and which is - depends on the detailed definition used for those vectors and the order in which they're taken for the calculations.)
answered Jan 25 at 4:01
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
You need the initial velocity of the asteroid (relative to a frame reference centered on the planet). If the asteroid were initially at rest in that frame, since gravity acts in the direction to the center of mass of the planet, the asteroid would follow a direct collision trajectory (a free fall on the planet).
If you have the velocity vector for the asteroid in the reference frame centered at the planet, then sign of the third component of the vector product of the position and velocity vectors (taken as vectors in $mathbb R^3$) predicts the orientation of the orbit. If you want to keep it two dimensional, think of the determinant of the matrix whose columns are the position and the velocity... again, the sign defines the orientation (which orientation is + and which is - depends on the detailed definition used for those vectors and the order in which they're taken for the calculations.)
answered Jan 25 at 4:01
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
You need the initial velocity of the asteroid (relative to a frame reference centered on the planet). If the asteroid were initially at rest in that frame, since gravity acts in the direction to the center of mass of the planet, the asteroid would follow a direct collision trajectory (a free fall on the planet).
If you have the velocity vector for the asteroid in the reference frame centered at the planet, then sign of the third component of the vector product of the position and velocity vectors (taken as vectors in $mathbb R^3$) predicts the orientation of the orbit. If you want to keep it two dimensional, think of the determinant of the matrix whose columns are the position and the velocity... again, the sign defines the orientation (which orientation is + and which is - depends on the detailed definition used for those vectors and the order in which they're taken for the calculations.)
answered Jan 25 at 4:01
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
You need the initial velocity of the asteroid (relative to a frame reference centered on the planet). If the asteroid were initially at rest in that frame, since gravity acts in the direction to the center of mass of the planet, the asteroid would follow a direct collision trajectory (a free fall on the planet).
If you have the velocity vector for the asteroid in the reference frame centered at the planet, then sign of the third component of the vector product of the position and velocity vectors (taken as vectors in $mathbb R^3$) predicts the orientation of the orbit. If you want to keep it two dimensional, think of the determinant of the matrix whose columns are the position and the velocity... again, the sign defines the orientation (which orientation is + and which is - depends on the detailed definition used for those vectors and the order in which they're taken for the calculations.)
answered Jan 25 at 4:01
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
edited Jan 25 at 4:08
answered Jan 25 at 4:01
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Jan 25 at 4:01
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Jan 25 at 4:01
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
4,765118
add a comment |
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1
$begingroup$
The vector product?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 4:00
$begingroup$
Can you elaborate more?
$endgroup$
– Gary Holiday
Jan 25 at 4:01
$begingroup$
What's point 1, magnitude of asteroid supposed to mean?
$endgroup$
– Ankit Kumar
Jan 25 at 4:21