Question

According to Kepler's Second Law the radius vector drawn from the Sun to a planet Multiple Choice is the same for all planets. sweeps out equal areas in equal times. sweeps out a larger area for a given time when the planet is moving faster. sweeps out a larger area for a given time when the distance to the Sun is greater.

Answer 1

sweeps out equal areas in equal times.

Step-by-step explanation:

As we know that there is no torque due to Sun on the planets revolving about the sun

so we will have

`\tau_(net) = 0`

now we have

`(dL)/(dt)= 0`

now we also know that

`Area = (1)/(2)r^2d\theta`

so rate of change in area is given as

`(dA)/(dt) = (1)/(2)r^2(d\theta)/(dt)`

so we will have

`(dA)/(dt) = (1)/(2)r^2\omega`

`(dA)/(dt) = (L)/(2m)`

since angular momentum and mass is constant here so

all planets sweeps out equal areas in equal times.