Suppose Halley's comet orbits the Sun every 74 years and has an eccentricity of 0.94. What is the perihelion distance of Halley’s comet in AU?

Answer 1

We can start with an equation relating period of an elliptical orbit to its semi-major axis:

Period (T)² = 4π²(semi-major axis (a))³ / gravitational constant (G)(mass of the sun (M) + mass of the comet (m))

m is so much smaller than M that we can ignore it

(74 yr * 365 d/yr * 24 h/d * 3600 s/h)² = 4π²a³ / (6.67 x 10^-11)(2 x 10³⁰)

a = 2.64 x 10¹² m * 1 astronomical unit (AU)/ 1.5 x 10¹¹ m = 17.6 AU

In an ellipse, perihelion distance = a - a(eccentricity) = 17.6 -17.6(0.94) = 1.06 AU