A small body starts falling onto the Sun from a distance equal to the radius of the Earth's orbit. The initial velocity of the body is equal to zero in the heliocentric reference frame. Making use of Kepler's laws, how long the body will be falling in days

64.5

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65

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60

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None of these

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Solution

The correct option is A $64.5$

We can think of the body as moving in a very elongated orbit of maximum distance $R$ and minimum distance 0 so semi major axis $= \frac{R}{2}$ . Hence if $τ$ is the time of fall then

${(\frac{2 τ}{T})}^{2} = {⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{R}{2}}{R} ⎞ ⎟ ⎟ ⎟ ⎠}^{3}$ or $τ^{2} = \frac{T^{2}}{32}$

or $τ = \frac{T}{4 \sqrt{2}} = \frac{365}{4 \sqrt{2}} = 64.5 d a y s$ .

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A small body starts falling onto the Sun from a distance equal to the radius of the Earth's orbit. The initial velocity of the body is equal to zero in the heliocentric reference frame. Making use of Kepler's laws, how long the body will be falling in days

The correct option is A 64.5 We can think of the body as moving in a very elongated orbit of maximum distance R and minimum distance 0 so semi major axis =R2. Hence if τ is the time of fall then (2τT)2=⎛⎜ ⎜ ⎜⎝R2R⎞⎟ ⎟ ⎟⎠3 or τ2=T232 or τ=T4√2=3654√2=64.5days.