Respuesta :
F = m A
A = Ac = v^2/R
so m A = Mjupiter v^2/R (toward sun)
F = G Msun MJupiter /R^2 (toward sun)
so
G Msun/R^2 = v^2/R
G Msun = v^2 R
Time around = circumference /v
T = 2 pi R/v
so
v = 2 pi R/T
v^2 = (2pi)^2 R^2/T^2
so
G Msun = (2 pi)^2 R^3/T^2
(which by the way is Kepler's Third Law)
so
R^3 = G Msun T^2/(2 pi)^2
G is 6.67*10^-11
so
R^3 = 6.67*10^-11*1.99*10^30*14.36*10^16 /39.48
R^3 = 4.828*10^35
= .4828 * 10^36
so
R = .784 * 10^12 = 7.84 * 10^11 meters
A = Ac = v^2/R
so m A = Mjupiter v^2/R (toward sun)
F = G Msun MJupiter /R^2 (toward sun)
so
G Msun/R^2 = v^2/R
G Msun = v^2 R
Time around = circumference /v
T = 2 pi R/v
so
v = 2 pi R/T
v^2 = (2pi)^2 R^2/T^2
so
G Msun = (2 pi)^2 R^3/T^2
(which by the way is Kepler's Third Law)
so
R^3 = G Msun T^2/(2 pi)^2
G is 6.67*10^-11
so
R^3 = 6.67*10^-11*1.99*10^30*14.36*10^16 /39.48
R^3 = 4.828*10^35
= .4828 * 10^36
so
R = .784 * 10^12 = 7.84 * 10^11 meters
The mean distance between the center of the Jupiter and the center of the Sun is "7.85 x 10¹¹ m"
The force of gravitation between the Sun and Jupiter must be equal to the centripetal force between them, for the equilibrium revolution of Jupiter around the Sun.
[tex]Centripeta\ Force\ on\ Jupiter = Gravitational\ Force\ of Attraction\ \\\\\frac{M_{Jupiter}v^2}{r} = \frac{GM_{Jupiter}M_{Sun}}{r^2}\\\\v^2 = \frac{GM_{Sun}}{r}\ -------- eqn(1)\\\\[/tex]
where,
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
[tex]M_{Sun}[/tex] = Mass of Sun = 1.99 x 10³⁰ kg
r = mean distance between the center of the Jupiter and the Sun = ?
v = linear speed of the Jupiter around the Sun = [tex]\frac{Circumference\ of Jupiter's\ Path}{Time\ Period\ of\ Revolution}[/tex]
[tex]v = \frac{2\pi r}{3.79\ x\ 10^8\ s}\\\\v^2 = \frac{4\pi^2 r^2}{14.36\ x\ 10^{16}\ s^2}[/tex]
Using the values in eqn (1), we get:
[tex]\frac{4\pi^2 r^2}{14.36\ x\ 10^{16}\ s^2} = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{r}\\\\r^3 = \frac{(14.36\ x\ 10^{16}\ s^2)(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{4\pi^2}\\\\r = \sqrt[3]{4.83\ x\ 10^{35}\ m^3}[/tex]
r = 7.85 x 10¹¹ m
Learn more about centripetal force and gravitational force here:
https://brainly.com/question/14021112?referrer=searchResults
https://brainly.com/question/16613634?referrer=searchResults
The attached picture shows the relationship between the centripetal force and the gravitational force acting on a planet (Jupiter) revolving around the sun.