# Tag Info

The Earth is currently rotating at one revolution per sidereal day. Converting to radians, this is an angular velocity of $\omega_0 = \frac {2\pi}{\text{sidereal day}}$. You want to have 360 solar days per year, or 361 sidereal days per year. That means a rotation rate of one revolution per 1/361 tropical year, or an angular velocity of $\omega_1 = \frac ... 2 Consider a rod held vertically with a pivot at its base. If the rod is allowed to fall to one side each point on the rod turns through the same angle in a given amount of time. This is shown in the diagram bellow: To say otherwise would be against all experience. The definition of angular acceleration is the rate of change of angular velocity or the ... 2 How fast would a sphere need to rotate for a dust speck at its equator to achieve balance between gravitational attraction and centrifugal force? If you do the math (equating$G M m / R^2$to$m \omega^2 R$and using$M = \frac{4\pi}{3} \rho R^3$as well as$\omega = 2\pi f$), it follows that the size of the sphere is entirely irrelevant and that only the ... 1 Original answer This question when applied to the Earth is purely academic. The easiest solution is the one posted by Johannes: One revolution per one hour and 24 minutes. Why go beyond that? The question is academic. It's not academic when applied to asteroids. There's an interesting effect, the Yarkovsky–O'Keefe–Radzievskii–Paddack effect, aka the YORP ... 1 I don't want to give away the answer immediately, since it's homework. Try to visualize one huge big wheel and one very small wheel. Try to imagine what would happen if at each instant the tangential velocity is not the same. 1 No. Kepler's second law says that "A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time." This is equivalent to saying that the "angular velocity" is inversely proportional to the distance relative to the centre of mass. In other words it slows down as it gets further away in the orbit. If the angular speed were ... 1 Ignoring the (minor) effects due to the other planets, the angular momentum of the Earth-Sun system must be conserved, and the angular momentum is given by (making the approximation that the Sun is fixed): $$L = \omega m_e r_e^2$$ where$m_e$is the mass of the Earth and$r_e^2\$ is the Earth-Sun distance. A quick rearrangement to get the formula for the ...