is it possible to find the orbital velocity of the earth around sun?

Question

i want to have daily velocity of the earth around sun for 2001 with concidering that the earth's speed for each day is different

Answer 1

The Earth moves faster in its orbit as it gets closer to the Sun and slower as it moves away. The speed of Earth's orbit increases roughly by 1% as it gets 1% closer to the Sun and decreases in speed by 1% as it gets 1% further from the Sun.

The Earth travels at about 18.5386 miles per second around Jan 4th when the Earth is about 91,652,250.885 miles away from the Sun (perihelion).

The Earth travels at about 18.4758 miles per second around July 4th when the Earth is about 94,821,243.935 miles away from the Sun (aphelion).

The difference in miles between farthest and closest is 3,168,993.05 miles.

The difference in speed between farthest and closest is 0.0628 miles per second.

Knowing a 1% increase or decrease in speed requires a 1% increase or decrease in distance from the Sun, we can calculate 1% of distance from 3,168,993.05 miles, which is 31,689.9305 miles. So, for every 31,689.9305 miles the Earth moves closer or further away from the Sun, the speed will increase/decrease by 1%. Being that 0.0628 miles per second is the total difference in speed change (100%), we can calculate 1% of that to be 0.000628 miles per second.

On Jan 4th at 5 am EST the Earth was at perihelion (closest to the Sun and fastest in speed). At a top speed of 18.5386 miles per second, we divide 31,689.9305 by 18.5386 to get 1709.4 secs (or 28.49 hours) for the Earth to travel 31,689.9305 miles further away from the Sun. Add 28.49 hours to 5 AM EST on Jan 4 and we come to around 9:30 AM EST on Jan 5. That's when we can decrease the speed by 0.000628 miles per second (since it is now starting to move away from the Sun), which would put the Earth's speed at 18.537972 miles per second after 9:30 AM EST on Jan 5.

We would then do the above calculation over again with the new speed to know when we can decrease the speed by 1% again.
31,689.9305 / 18.537972 = 1709+/- seconds = 28.491 hours from 9:30 AM EST on Jan 5 = 2 PM EST Jan 6 (roughly). Then we can decrease speed by 1% again (0.000628 miles per second ) and recalculate with the new speed to figure out how long it will take Earth to travel another 31,689.9305 miles away from the Sun. This will give us the numbers we need to calculate the time and date when we can reduce speed by 1% once again.

Once the Earth reaches aphelion around July 4th (furthest from the Sun and slowest in speed), we then start to increase the speed by 1% each time until we get back to perihelion again, which should be sometime around the following Jan 4th.

Answer 2

If you know the semi-major axis and the date that the Earth is at perigee. You can find the perigee date in an astronomical almanac - it's usually in the first week of January. You can calculate the daily speed using several formulas:

Speed:
v = sqrt[ GM * (2/r - 1/a)]
GM is the universal gravitational constant times the mass of the Sun. You can look up the semi-major axis in an astronomical almanac. You have to calculate the radius.

Radius for a given day:
r= [a(1-e^2)]/[1-e*cos(nu)]
You can look up the eccentricity in an astronomical almanac. nu is the true anomaly. You have to calculate it.

cos(nu) = [cos(E) - e]/[1 - e* cos(E)]
E is the eccentric anomaly. You have to calculate it from Mean Anomaly (M).

M = E - e * sin(E)
There is no exact solution for finding E from M. You can find a very close approximation using a trial and error method, such as Newton's method.

Mean Anomaly could be considered an orbiting object's location with respect to time. Divide 2pi by the orbital period (in days) and you have the average angular velocity (mean motion). Multiply the mean motion by the number of days since perigee to get Mean Anomaly.

Answer 3

at first, there are forces that could reason the earth's orbit to decay. the main glaring of those are friction (regardless of what you assert) and tidal forces. Friction could be assume to happen between the image voltaic wind and the earth. "Empty" area isn't - not extremely. To the quantity that the image voltaic wind consists of electrically impartial debris, those trave in straight away strains faraway from the sunlight, and the earth is often working into them at its maximum advantageous hemisphere. the different rigidity is exactly the comparable using fact the rigidity which reasons the moon's orbit to decay. The sunlight will improve tides in the international (even if image voltaic tides are smaller than lunar tides),a nd in precisely the comparable way, the earth reasons tides on the sunlight. the result's that the gravitational rigidity between the earth and the sunlight isn't precisely alongside the line connecting their centres of gravity. The power of the earth's progression around the sunlight and of its rotation approximately its axis is became, in a modest way, into frictional warmth. even nonetheless, the two those effects are so very, very small which you would be able to forget approximately approximately them. they're insignificant on the timescale of the probably existence of the earth.

Answer 4

The earth goes around the sun in one year, that's 31,556,926 seconds. The sun is 92,900,000 miles away on average. Multiply that by 2 and by pi, giving a circumference of 583.7 million miles. Divide that by 31,556,926. That makes 18.5 miles per second. By the way, escape velocity from the earth's orbit is its orbital velocity times the square root of 2, or 26 miles per second to escape the solar system.