Sun will turn into a red giant star one day. Today the sun has a mass of 2*10^30kg and a radius of 70km and an angular velocity of 2.85*10^-6 rad/s.
How fast are the particles moving at the surface of the sun? When the sun turns into a red giant its mass will be 75% of what it is today and its radius will be 80 times larger calculate red giants angular velocity.
tangential velocity= radius*angularvelocity.
70km is not very far, I can drive that in about 40 minutes. I think the Sun must be larger than that.
To calculate the speed of particles at the surface of the sun, we can use the formula for the tangential velocity of a rotating object:
v = R * ω
Where:
- v is the tangential velocity
- R is the radius
- ω is the angular velocity
Given:
- R = 70 km = 70,000 m
- ω = 2.85 * 10^-6 rad/s
Substituting these values into the formula, we can find the tangential velocity of particles at the surface of the sun:
v = (70,000 m) * (2.85 * 10^-6 rad/s)
v ≈ 0.1995 m/s
Therefore, the particles at the surface of the sun are moving at approximately 0.1995 meters per second.
To calculate the red giant's angular velocity, we need to take into account the change in mass and radius. Given that the red giant's mass will be 75% of its current mass and its radius will increase by a factor of 80, we can use the principle of conservation of angular momentum:
I * ω = I' * ω'
Where:
- I is the moment of inertia of the sun (mass * radius^2)
- ω is the current angular velocity
- I' is the moment of inertia of the red giant
- ω' is the angular velocity of the red giant
We know that I = (2 * 10^30 kg) * (70,000 m)^2
And I' = (0.75 * 2 * 10^30 kg) * (80 * 70,000 m)^2
We can rearrange the equation to solve for ω':
ω' = (I * ω) / I'
Substituting the values we know:
ω' = [(2 * 10^30 kg) * (70,000 m)^2 * (2.85 * 10^-6 rad/s)] / [(0.75 * 2 * 10^30 kg) * (80 * 70,000 m)^2]
After solving this equation, we find the angular velocity of the red giant.