Log in to add a comment

## Answers

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

Keplers 2nd law says that a planet revolving
around the Sun at the focus of the elliptical orbit, sweeps the same areas in
the same period of time. We want to prove that.

The area ΔA swept (in time Δt) by a planet of mass m moving with instantaneous velocity v at a instantaneous distance r from the Sun,

ΔA = 1/2 (r Χ v Δt ) as the Δs = arc length = v Δt

the quantities in bold are vectors.

ΔA/Δt = 1/2 (r Χ v) = 1/2 (r Χ p/m) = L / 2m

L is the angular momentum vector of the planet.

As there is not external force on the planet and Sun system, the total angular momentum is a constant. Hence L is constant. That means ΔA/Δt is constant.

Hence Area is directly proportional to time or Area swept is constant for a constant duration of time.

The area ΔA swept (in time Δt) by a planet of mass m moving with instantaneous velocity v at a instantaneous distance r from the Sun,

ΔA = 1/2 (r Χ v Δt ) as the Δs = arc length = v Δt

the quantities in bold are vectors.

ΔA/Δt = 1/2 (r Χ v) = 1/2 (r Χ p/m) = L / 2m

L is the angular momentum vector of the planet.

As there is not external force on the planet and Sun system, the total angular momentum is a constant. Hence L is constant. That means ΔA/Δt is constant.

Hence Area is directly proportional to time or Area swept is constant for a constant duration of time.