Respuesta :
Answer:
Correct Answer is E(7.8\times10^{11} meters[/tex]
)
Explanation:
In this question we have given,
Time period= 12 years ([tex]3.79\times10^{8}seconds[/tex])
mass of the Sun, M = [tex]1.99\times10^{30}kg[/tex]
Let m be the mass of jupiter
We know that,
[tex]Force =mass\times acceleration[/tex]
[tex]F =m\times a[/tex].............(1)
Also we know that,
[tex]a=\frac{v^2}{R}[/tex]
put values of m(mass of jupiter) and a in equation (1)
[tex]F =m\times \frac{v^2}{R}[/tex]..............(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
[tex]F=\frac{GmM}{R^2}[/tex] .............(3)(Direction of this force is toward sun)
Here G is universal gravitational constants and its value is [tex]6.67\times10^{-11}m^{3}kg^{-1}s^{-2}[/tex]
From equation (2) and equation (3)
[tex]m\times \frac{v^2}{R}=\frac{GmM}{R^2}[/tex]
[tex]\frac{v^2}{R}=\frac{GM}{R^2}[/tex]
[tex]GM= v^{2}R[/tex]...................(4)
We know that,
Time period [tex]= \frac{circumference}{v}[/tex]
[tex]T=\frac{2\pi R}{v}[/tex]
or
[tex]v=\frac{2\pi R}{T}[/tex]
put value of v in equation (4)
we get,
[tex]GM=\frac{(2\pi )^2R^3}{T^2}[/tex]
Therefore,
[tex]R^3=\frac{G MT^2}{(2\pi )^2}[/tex].............(5)
put values of G, M, T and [tex]\pi[/tex] in equation 5
[tex]R^3 =\frac{6.67\times10^{-11}m^{3}kg^{-1}s^{-2}\times1.99\times10^{30}kg\times14.36\times10^{16}s^{2} }{39.48}[/tex]
[tex]R^{3}=4.828\times10^{35}[/tex]
or,
[tex]R = 0.784\times10^{12}\\R = 7.8\times10^{11} meters[/tex]
(in round figure 7.84 can be written as 7.8)
Answer:
The mean distance from the center of the Sun is [tex]7.84\times 10^{11}\ m[/tex]
Explanation:
It is given that,
The planet Jupiter revolves around the Sun in a period of about 12 years, [tex]T=3.79\times 10^8\ seconds[/tex]
Mass of the sun, [tex]M=1.99\times 10^{30}\ kg[/tex]
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.
[tex]T^2\propto a^3[/tex]
a = mean distance from the center of the sun.
[tex]T^2=\dfrac{4\pi^2}{GM}a^3[/tex]
[tex]a^3=\dfrac{T^2GM}{4\pi^2}[/tex]
[tex]a^3=\dfrac{(3.79\times 10^8\ s)^2\times 6.67\times 10^{-11}\times 1.99\times 10^{30}\ kg}{4\pi^2}[/tex]
[tex]a=(4.82\times 10^{35})^{\dfrac{1}{3}}\ m[/tex]
[tex]a=7.84\times 10^{11}\ m[/tex]
So, the mean distance from the center of the sun is [tex]7.84\times 10^{11}\ m[/tex]. Hence, this is the required solution.
