Question

A comet has a very elliptical orbit with a period of 137.6 y. If the closest approach of the comet to the Sun is 0.09 AU, what is its greatest distance from the Sun

Answer 1

53.35 AU

Step-by-step explanation:

Using Kelper's third law

The relation between the time period (T) and the semi-major axis of the comet can be expressed as:

T²∝a³

T² = Ka³

where;

(a) = length of the semi-major axis

Making (a) the subject of the axis

a³
`=(T^2)/(K)`

`a = \sqrt[3]{(T^2)/(K) }`

`a = \sqrt[3]{(((137.6y)((3.16*10^7sec)/(1year))^2)/(2.96*10^(-19)s^2/m^3) }`

`a =(3.997*10^(12)m)((1AU)/(149597870700m) )`

a = 26.72 AU

Finally, the greatest distance from the Sun is:

`d = 2a -`
`d_(approach)`

d = 2(26.72) - 0.09

d =53.44 - 0.09

d = 53.35 AU

Thus, the greatest distance from the Sun = 53.35 AU