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.

Respuesta :

Answer:

sweeps out equal areas in equal times.

Explanation:

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

so we will have

[tex]\tau_{net} = 0[/tex]

now we have

[tex]\frac{dL}{dt}= 0[/tex]

now we also know that

[tex]Area = \frac{1}{2}r^2d\theta[/tex]

so rate of change in area is given as

[tex]\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt}[/tex]

so we will have

[tex]\frac{dA}{dt} = \frac{1}{2}r^2\omega[/tex]

[tex]\frac{dA}{dt} = \frac{L}{2m}[/tex]

since angular momentum and mass is constant here so

all planets sweeps out equal areas in equal times.

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