<p><span class="math-container">$d\phi/dt$</span> is an angular speed, so yes you should set this to be an angular speed. You say the coordinate system co-rotates with the magnetic field, so you should also apply any offset for this.</p>
<p>If the initial speed is 20 km/s, then the angular speed in radians/s is given by 20,000 divided by the radial coordinate in metres.</p>

CC BY-SA 4.0
2023-10-10T12:04:07.483Z
Thank you. That was simple and clear response. As regard to the coordinate, its centre is taken to be centre of the planet i.e. the magnetic field.
Lunthang Peter
CC BY-SA 4.0
2023-10-10T12:25:58.180Z
@LunthangPeter I assumed that. The issue is the co-rotating coordinate system. Your angular velocity should be with respect to the magnetic field, not with respect to a non-rotating frame.
ProfRob
CC BY-SA 4.0
2023-10-10T13:27:13.127Z
Yes. Point Noted.
Lunthang Peter
CC BY-SA 4.0
2023-10-09T11:19:33.220Z
2023-10-09T11:19:33.220Z
ProfRob
<p>A charged particle is moving in a planetary magnetic field, in a spherical polar coordinates which is corotating with the magnetic field. To solve the equations of motion numerically what would be the initial azimuthal speed <span class="math-container">$\frac{d\phi}{dt}$</span>, if the particle initially moving with the Kepler speed,<span class="math-container">$\sqrt{\frac{GM}{R}}$</span> along the equatorial in the co rotational direction? Do we need convert the keppler speed into the corresponding angular speed, <span class="math-container">$\sqrt{\frac{GM}{R^3}}$</span> as the azimuthal speed is in radian/s while the kepler speed is in m/s? The same question goes what if I only know the initial speed in km/s, say 20km/s how would I input this?</p>

<astrophysics><stellar-astrophysics><planetary-science><units><high-energy-astrophysics>
When analyzing a particle moving at Kepler speed, √(GM/R) in a planetary magnetic field, do we need to convert it to angular speed, √(GM/R^3)?
Lunthang Peter
# When analyzing a particle moving at Kepler speed, √(GM/R) in a planetary magnetic field, do we need to convert it to angular speed, √(GM/R^3)?

(Asked by: username_0 on 2023-10-08)

A charged particle is moving in a planetary magnetic field, in a spherical polar coordinates which is corotating with the magnetic field. To solve the equations of motion numerically what would be the initial azimuthal speed $\frac{d\phi}{dt}$, if the particle initially moving with the Kepler speed,$\sqrt{\frac{GM}{R}}$ along the equatorial in the co rotational direction? Do we need convert the keppler speed into the corresponding angular speed, $\sqrt{\frac{GM}{R^3}}$ as the azimuthal speed is in radian/s while the kepler speed is in m/s? The same question goes what if I only know the initial speed in km/s, say 20km/s how would I input this?

**Tags:** astrophysics, stellar-astrophysics, planetary-science, units, high-energy-astrophysics

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## 1 answers

### Answer by username_1 on 2023-10-09. Score: 2 | Accepted answer

$d\phi/dt$ is an angular speed, so yes you should set this to be an angular speed. You say the coordinate system co-rotates with the magnetic field, so you should also apply any offset for this.

If the initial speed is 20 km/s, then the angular speed in radians/s is given by 20,000 divided by the radial coordinate in metres.

> **Comments:**
> **username_0**: Thank you. That was simple and clear response. As regard to the coordinate, its centre is taken to be centre of the planet i.e. the magnetic field.  
> **username_1**: @LunthangPeter I assumed that. The issue is the co-rotating coordinate system. Your angular velocity should be with respect to the magnetic field, not with respect to a non-rotating frame.  
> **username_0**: Yes. Point Noted.  

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