Question

Jupiter's semimajor axis is 7.78×1011 m. The mass of the Sun is 1.99×1030 kg. (a) What is the period of Jupiter's orbit in seconds? (b) What is the period in Earth years? Assume that one Earth year is exactly 365 days, with 24 hours in each day.

Answer 1

Step-by-step explanation:

It is given that,

Semi major axis of the Jupiter,
`a=7.78* 10^(11)\ m`

Mass of the sun,
`M=1.99* 10^(30)\ kg`

(a) Let T is the period of Jupiter's orbit. It is given by :

`T^2\propto a^3`

`T^2=(4\pi^2)/(GM)a^3`

`T^2=(4\pi^2)/(6.67* 10^(-11)* 1.99* 10^(30))* (7.78* 10^(11))^3`

`T=3.74* 10^8\ s`

(b) We know that,

`1\ year=3.154* 10^7\ s`

or

`1\ s=3.171* 10^(-8)\ year`

`3.74* 10^8\ s={3.171* 10^(-8)}* {3.74* 10^8}`

T = 11.859 earth years

Hence, this is the required solution.

Answer 2

(a) 3.74 x 10^8 seconds

(b) 11.86 earth year

Step-by-step 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.

`T=2\pi \sqrt{(R^(3))/(GM)}`

`T=2\pi \sqrt{(\left ( 7.78*10^(11) \right )^(3))/(6.67*10^(-11)*1.99*10^(30))}`

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.