Respuesta :
Explanation:
It is given that,
Semi major axis of the Jupiter, [tex]a=7.78\times 10^{11}\ m[/tex]
Mass of the sun, [tex]M=1.99\times 10^{30}\ kg[/tex]
(a) Let T is the period of Jupiter's orbit. It is given by :
[tex]T^2\propto a^3[/tex]
[tex]T^2=\dfrac{4\pi^2}{GM}a^3[/tex]
[tex]T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.99\times 10^{30}}\times (7.78\times 10^{11})^3[/tex]
[tex]T=3.74\times 10^8\ s[/tex]
(b) We know that,
[tex]1\ year=3.154\times 10^7\ s[/tex]
or
[tex]1\ s=3.171\times 10^{-8}\ year[/tex]
[tex]3.74\times 10^8\ s={3.171\times 10^{-8}}\times {3.74\times 10^8}[/tex]
T = 11.859 earth years
Hence, this is the required solution.
Answer:
(a) 3.74 x 10^8 seconds
(b) 11.86 earth year
Explanation:
Radius, R = 7.78 x 10^11 m
mass of sun, M = 1.99 x 10^30 kg
(a) Let T be the period of Jupiter.
[tex]T=2\pi \sqrt{\frac{R^{3}}{GM}}[/tex]
[tex]T=2\pi \sqrt{\frac{\left ( 7.78\times10^{11} \right )^{3}}{6.67\times10^{-11}\times1.99\times10^{30}}}[/tex]
T = 3.74 x 10^8 seconds
(b) 1 year = 365 days
1 day = 24 hour
1 hour = 60 minutes
1 minute = 60 seconds
T = 11.86 earth year.
