Question

Kepler's second law states that a line from the sun to a planet sweeps out equal areas in equal times. 2 d coordinate system with an elliptical orbit of planet around the sun. The elliptical orbit is centered at the origin. The Sun is located on the x axis between the origin and the far left edge of the orbit. The Planet is on the orbital path in the first quadrant and has a velocity vector that is pointing counter-clockwise on the orbital path. There is a line connecting the sun and the planet. The Perihelion is marked as the point where the orbit crosses the x axis closest to the Sun. The Aphelion is marked as the point where the orbit crosses the x axis farthest away from the Sun. The aphelion is the point in the orbit where the planet is the furthest distance from the sun, The perihelion is the point at which the planet is closest to the sun. At which point must the speed of the planet be greater

Answer 1

the speed is higher in the PERIHELIUM

Step-by-step explanation:

As stated in your statement, Kepler's second law says that a vector from the sun to the planet also sweeps equal in equal times. This is a consequence that the sun-planet system is isolated, therefore the angular momentum is conserved.

L = r x p = m r x v

where m is the planet mass and the Sun is considered fixed

Let's analyze this expression, if the anglar momentum is a constant when r is less (perihelion) the speed must increase, so that the product remains fixed

So the speed is higher in the PERIHELIUM