A comet at its point of closest approach to the sun where the radial distance from the centre of the sun is $$50\times10^6$$ miles, has a velocity of 188,500 ft/sec. Determine the radial and transverse components of its velocity at a point B, where the radial distance from the centroid of the sun is $$75\times10^6$$ miles.

I've managed to determine that the comet will have a velocity of 23.80 mi/s at point B, but can someone explain how I would get the radial and transverse components of this? Thanks.

• What principle / logic did you use to get 23.80 mi/s ?
– AJN
Aug 19, 2022 at 16:37
• Inverse square potential fields have a hypersymmetry associated with them that introduces an extra conservative function. The one you are interested in is called the Laplace-Runge-Lentz vector. Aug 19, 2022 at 16:45
• The comet is not orbiting around the sun, without knowing the internal angle between the lines (from the sun to the points), there is no solution.
– r13
Aug 19, 2022 at 21:25

You could calculate the velocity magnitude from total energy (gravitational and kinetic) of the comet, which must be constant: $$E = E_p+E_k = -\frac{G\cdot M\cdot m}{r_A}+\frac{1}{2}m\cdot v_A^2 = -\frac{G\cdot M\cdot m}{r_B}+\frac{1}{2}m\cdot v_B^2$$ From that, the velocity magnitude $$v_B$$ at point B is: $$v_B = \sqrt{v_A^2-2\cdot G\cdot M\cdot \frac{r_B-r_A}{r_A\cdot r_B}}$$ (You will need gravitational constant $$G$$ and mass of the Sun $$M$$.)
For the velocity components, you can use second Keplers law, which states that area velocity $$\vec{A}$$ is constant: $$\vec{A} = \vec{r}_A \times \vec{v}_A = \vec{r}_B \times \vec{v}_B$$ From that, area velocity magnitude can be expressed as: $$A = r_A\cdot v_A \cdot \sin(\alpha) = r_B\cdot v_B \cdot \sin(\beta)$$ Since $$\alpha$$ is 90°, the angle $$\beta$$ between position vector $$\vec{r}_B$$ and velocity vector $$\vec{v}_B$$ will be: $$\beta = \arcsin\left(\frac{r_A\cdot v_A}{r_B\cdot v_B}\right)$$ Now the radial and transverse velocity components at point B are: $$\vec{v}_B = \left(v_B\cdot \cos(\beta), v_B\cdot \sin(\beta)\right)$$
• Good argument. Two comments. Firstly, the reason we know that $\alpha$ is a right angle is that "closest approach" means an extremum of radial distance, so the radial component of velocity there must be zero. Secondly, Kepler's second law is just a prettyfied geometric way of (p)restating conservation of angular momentum. Aug 21, 2022 at 13:55