Answer:
53.35 AU
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 [tex]=\frac{T^2}{K}[/tex]
[tex]a = \sqrt[3]{\frac{T^2}{K} }[/tex]
[tex]a = \sqrt[3]{\frac{((137.6y)(\frac{3.16*10^7sec}{1year})^2}{2.96*10^{-19}s^2/m^3} }[/tex]
[tex]a =(3.997*10^{12}m)(\frac{1AU}{149597870700m} )[/tex]
a = 26.72 AU
Finally, the greatest distance from the Sun is:
[tex]d = 2a -[/tex] [tex]d_{approach}[/tex]
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
