Post by telemeter on Mar 4, 2019 14:42:41 GMT
As Vasco says, in a central force field transverse acceleration is zero. Therefore we can equate the transverse acceleration term to zero and rearrange, collecting like terms on either side. The numerators are the derivative forms of the denominators so we can directly integrate both sides wrt t. This gives:
$2 \ln r = - \ln (\dot{\theta}) + \ln k$ , $k$ being a constant
So, $r^2= k/ (\dot{\theta})$ and hence $r^2 \dot{\theta} = k$
Interestingly (for me anyway), is that the angular momentum of a particle in such a field with transverse velocity, vt, is vt * r. Now vt = r * theta dot, so if one wants angular momentum to be conserved:
vt * r = constant = r * theta dot * r = r2 * theta dot.
So the conservation of angular momentum requires the constancy of areal speed. As this is at heart a physics exercise perhaps then that is a third (slightly off-topic admittedly) way to arrive at the required result.
A little thing, but I had never related the two before for some reason.
$2 \ln r = - \ln (\dot{\theta}) + \ln k$ , $k$ being a constant
So, $r^2= k/ (\dot{\theta})$ and hence $r^2 \dot{\theta} = k$
Interestingly (for me anyway), is that the angular momentum of a particle in such a field with transverse velocity, vt, is vt * r. Now vt = r * theta dot, so if one wants angular momentum to be conserved:
vt * r = constant = r * theta dot * r = r2 * theta dot.
So the conservation of angular momentum requires the constancy of areal speed. As this is at heart a physics exercise perhaps then that is a third (slightly off-topic admittedly) way to arrive at the required result.
A little thing, but I had never related the two before for some reason.
