Physics

Problem

An asteroid moves around the Sun in an elliptical orbit with semi-major axis

a = 4.00\,\text{AU}

and eccentricity

e = 0.50

. The orbital period is

8.00\,\text{years}

. The perihelion distance is therefore

2.00\,\text{AU}

and the aphelion distance is

6.00\,\text{AU}

. Assume that, during short intervals, the swept area can be taken as

\frac12 r^{2}\Delta\theta

, where

\Delta\theta

is in radians. Answer the following. a. (i) Calculate the areal velocity of the asteroid in

\text{AU}^2\,\text{year}^{-1}

. b. (ii) Find the time taken (in days) for the asteroid to advance through an angular displacement of

10^{\circ}

when it is near perihelion (

r = 2.00\,\text{AU}

). c. (iii) Find the corresponding time taken for the same

10^{\circ}

displacement when the asteroid is near aphelion (

r = 6.00\,\text{AU}

). d. (iv) Comment briefly on how the results illustrate the variation of orbital speed with distance from the Sun.

Intuitive explanation

Kepler's second law—the equal-area rule—tells us that an orbiting body sweeps out equal areas in equal times, which means the asteroid moves faster near perihelion (closest to the Sun) and slower near aphelion (farthest away). By combining this constant areal velocity with the geometry of the ellipse and the given orbital period, you can find the instantaneous speed and angular velocity at any point, or determine how long the asteroid spends in different regions of its orbit.

Intuitive explanation

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