Answer :
By equating the centripetal force and the gravitational pull, we have the equation
[tex]\frac{mv^2}{r}=\frac{GMm}{r^2}[/tex]
where
G=gravitational constant=6.67408*10^(-11) (m^3 kg^-1 s^-2)
m=mass of object
M=mass of sun=1.9885*10^(30) (kg)
v=tangential velocity of object
r=distance from the sun (m)
On simplification, we get the relation
[tex]v=\sqrt{\frac{GM}{r}}[/tex] (m/s)
Subtituting constants,including
1 au = 1.49597870700*10^(11) m
and given
r=5 au = 7.47989*10^11 m
[tex]v=\sqrt{\frac{GM}{r}}[/tex]
[tex]=\sqrt{\frac{1.32712*10^{20}}{7.47989*10^{11}}}[/tex]
[tex]=\sqrt{1.774250*10^8}[/tex]
[tex]=13320.10[/tex] m/s
Therefore the orbital period is
[tex]T=\frac{2\pi(5*au)}{13320.10}[/tex] s
[tex]=\frac{2\pi(5*au)}{13320.10*365.25*86400}[/tex] years
[tex]=11.181[/tex] years
[tex]\frac{mv^2}{r}=\frac{GMm}{r^2}[/tex]
where
G=gravitational constant=6.67408*10^(-11) (m^3 kg^-1 s^-2)
m=mass of object
M=mass of sun=1.9885*10^(30) (kg)
v=tangential velocity of object
r=distance from the sun (m)
On simplification, we get the relation
[tex]v=\sqrt{\frac{GM}{r}}[/tex] (m/s)
Subtituting constants,including
1 au = 1.49597870700*10^(11) m
and given
r=5 au = 7.47989*10^11 m
[tex]v=\sqrt{\frac{GM}{r}}[/tex]
[tex]=\sqrt{\frac{1.32712*10^{20}}{7.47989*10^{11}}}[/tex]
[tex]=\sqrt{1.774250*10^8}[/tex]
[tex]=13320.10[/tex] m/s
Therefore the orbital period is
[tex]T=\frac{2\pi(5*au)}{13320.10}[/tex] s
[tex]=\frac{2\pi(5*au)}{13320.10*365.25*86400}[/tex] years
[tex]=11.181[/tex] years