Question

When a planet is at its slowest orbital speed, its radius vector sweeps an area, A, in 45 days. What area will the radius vector for this planet sweep during a 40-day time period while at its fastest orbital speed

Answer 1

The area the radius vector will sweep is 0.889A

According to Kepler's second law, the radius vector sweeps out equal areas in equal times.

Let A = area and t = time period,

According to Kepler's law, A/t = constant

So, A₁/t₁ = A₂/t₂ where A₁ = area the radius vector sweeps at slowest orbital speed = A, t₁ = time period at slowest orbital speed = 45 days, A₂ = area the radius vector sweeps at fastest orbital speed, t₂ = time period at fastest orbital speed = 40 days.

Making A₂ subject of the formula, we have

A₂ = A₁t₂/t₁

Substituting the values of the variables into the equation, we have

A₂ = A₁t₂/t₁

A₂ = A × 40 days/45 days

A₂ = A × 40/45

A₂ = A × 8/9

A₂ = A × 0.889

A₂ = 0.889A

So, the area the radius vector will sweep is 0.889A

learn more about Kepler's second law here:

brainly.com/question/4639131