The sun has a mass of 1.99*10'30 kg, and is moving in a circular orbit around the center of our galaxy, the Milky way, dragging the rest of the solar system behing it. The radius of the orbit is 2.3*10'4 light years(1 ly=9.5*10'15)and the sun has an angular speed of 1.1*10'-15 rad/s. What is the magnitude of the net force keeping the sun moving on this path?
2007-11-30 09:52:26 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
F = m w^2 R = 1.99*10^30*(1.1*10^-15)^2*2.3*9.5*10^19 =
= 5.26*10^20 N
2007-11-30 10:16:23 · answer #1 · answered by Luigi 74 7 · 0⤊ 0⤋
The angular velocity of the sun is given and the radius of the orbit is given, so the angular acceleration is found using
a = r Ï^2
where you must first convert radius from light-years into meters. The conversion is given.
You are also given the mass of the sun, so the force on the Sun is then found from Newton's Second Law
F = ma
2007-11-30 18:58:54 · answer #2 · answered by jgoulden 7 · 0⤊ 0⤋
This sounds like a question for Einstein to answer.
So, according Einstein's general theory of relativity, the gravitational force on the sun would be Zero, because gravity is not a force - at least not in the theory of relativity.
On the other hand, if you wanted to divert the sun from its path around the galaxy...that would take a tremendous amount of force.
2007-11-30 18:05:29 · answer #3 · answered by farwallronny 6 · 0⤊ 0⤋