Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi). Ignoring the influence of other solar system bodies, Earth's orbit is an ellipse with the Earth-Sun barycenter as one focus and a current eccentricity of 0.0167; since this value is close to zero, the center of the orbit is close, relative to the size of the orbit, to the center of the Sun.
As seen from Earth, the planet's orbital prograde motion makes the Sun appear to move with respect to other stars at a rate of about 1° eastward per solar day (or a Sun or Moon diameter every 12 hours). Earth's orbital speed averages 29.78 km/s (107,208 km/h; 66,616 mph), which is fast enough to cover the planet's diameter in 7 minutes and the distance to the Moon in 4 hours.From a vantage point above the north pole of either the Sun or Earth, Earth would appear to revolve in a counterclockwise direction around the Sun. From the same vantage point, both the Earth and the Sun would appear to rotate also in a counterclockwise direction about their respective axes.
Hey everyone. I'm kind of new to the whole physics world, but I've got a question I've been wondering about.
So the Sun attracts the Earth because of gravity (or bends space time that the Earth is traveling through), and my question is this: Since photons have momentum, and momentum must be...
Given the mass of the sun, the gravitational constant, the period of Earth's orbit, and the semi-major axis of Earth's orbit, is it possible to find the eccentricity of the orbit? If yes, how?
Hi there, I've got a theoretical question i was hoping to gauge opinion about:
Is there any feasible situation which could change the Earth's orbit (without obliterating all life in the process)? And if so what would you imagine the effects to be, say for example if the Earth were to move...
Hi,
If we were to change the Earth's orbit, what force should we apply and in what direction? Shoud we go against G.(M_Earth + M_Sun) / distance^2 or against the centripetal force, M_Earth.v^2/distance?
I've calculated this last one and I got the result:
3,542396634E+22 Newtons
Is...
Yes i understand that i posted this earlier, but it was perhaps just glanced over since it had got replies (by me trying to bump it up, sorry, again). But i seriously would like to know how to solve this question ANy help would be appreciated
Note that i have copiedd the question exactly from...
i need help on this.
how much greater is the angular momentum of the Earth orbiting about the sun than the moon orbiting about the earth? (using a ratio of angular momenta)
angular momentum = rotational inertia x rotational velocity
radius of Earth (equatorial) 6.37x10^6
radius of...
kinetic energy...
The Earth's distance from the sun varies from 1.474e8 km to 1.525e8 km during the year. Take the Sun to be at rest.
Determine the difference in the Earth's kinetic energy.
Would I just use K=.5mv^2? If so what would v be?
If not, how would I set this up?
Thank you.