Question

The planet Jupiter revolves around the Sun in a period of about 12 years (3.79 × 108 seconds). What is its mean distance from the center of the Sun? The mass of the Sun is 1.99 × 1030 kilograms.

A. 1.1 × 1011 meters
B. 1.5 × 1011 meters
C. 2.3 × 1011 meters
D. 5.8 × 1011 meters
E. 7.8 × 1011 meters

Answer 1

The mean distance from the center of the Sun is
`7.84* 10^(11)\ m`

Step-by-step explanation:

It is given that,

The planet Jupiter revolves around the Sun in a period of about 12 years,
`T=3.79* 10^8\ seconds`

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

We need to find the mean distance from the center of the Sun. It can be calculated using the Kepler's third law of planetary motion i.e.

`T^2\propto a^3`

a = mean distance from the center of the sun.

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

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

`a^3=((3.79* 10^8\ s)^2* 6.67* 10^(-11)* 1.99* 10^(30)\ kg)/(4\pi^2)`

`a=(4.82* 10^(35))^{(1)/(3)}\ m`

`a=7.84* 10^(11)\ m`

So, the mean distance from the center of the sun is
`7.84* 10^(11)\ m`. Hence, this is the required solution.

Answer 2

Correct Answer is E(7.8\times10^{11} meters[/tex]

)

Step-by-step explanation:

In this question we have given,

Time period= 12 years (
`3.79*10^(8)seconds`)

mass of the Sun, M =
`1.99*10^(30)kg`

Let m be the mass of jupiter

We know that,

`Force =mass* acceleration`

`F =m* a`.............(1)

Also we know that,

`a=(v^2)/(R)`

put values of m(mass of jupiter) and a in equation (1)

`F =m* (v^2)/(R)`..............(2) (Direction of this force is toward sun)

We know that Gravitational force between jupiter of mass m and sun of Mass M which are at distance R is given as

`F=(GmM)/(R^2)` .............(3)(Direction of this force is toward sun)

Here G is universal gravitational constants and its value is
`6.67*10^(-11)m^(3)kg^(-1)s^(-2)`

From equation (2) and equation (3)

`m* (v^2)/(R)=(GmM)/(R^2)`

`(v^2)/(R)=(GM)/(R^2)`

`GM= v^(2)R`...................(4)

We know that,

Time period
`= (circumference)/(v)`

`T=(2\pi R)/(v)`

or

`v=(2\pi R)/(T)`

put value of v in equation (4)

we get,

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

Therefore,

`R^3=(G MT^2)/((2\pi )^2)`.............(5)

put values of G, M, T and
`\pi` in equation 5

`R^3 =(6.67*10^(-11)m^(3)kg^(-1)s^(-2)*1.99*10^(30)kg*14.36*10^(16)s^(2) )/(39.48)`

`R^(3)=4.828*10^(35)`

or,

`R = 0.784*10^(12)\\R = 7.8*10^(11) meters`

(in round figure 7.84 can be written as 7.8)