How to calculate the area of the mercury's orbit?

Question

The orbit of the mercury likes a section of egg.

Answer 1

It's so close to circular that you can just assume it to be a circle. Find the radius of its orbit (distance from mercury to sun), multiply it by itself, then multiply by Pi.

Answer 2

You are Johannes Kepler.  It is the year 1608 and you have successfully determined that the orbit of Mars is an ellipse.  Before publishing your results for the world to see, you want to determine the orbit of at least one other planet.  It may be a little risky to state that all planets move in elliptical orbits since you have determined this for only one planet out of five.  So you choose Mercury as your next object for study.  If the orbit of this planet is not a circle, but an ellipse, you will feel justified in generalizing to all planets and become famous for your discovery.  Just think, "Kepler's First Law."

You have a sheet of paper with an accurate scale drawing of the earth's orbit.  The dates given on the orbit are the earth's orbit.  The dates given on the orbit are the earth's positions on those dates, which are marked of at ten day intervals.  However since the drawing must be very accurate to get the correct orbital shape, you must interpolate to the exact dates given in the table of greatest elongations.  This can be done with a ruler.



Procedure (All work should be done in pencil.)

1. Determine the exact position of the earth in its orbit for the first greatest elongation.

2. Starting at this determined position draw a tangent line approximately 2/3 of the way across the circle at the appropriate angle from the sun's direction. (Use the protractors supplied by your instructor and in the way described by your instructor to get the angle accurately, Use a ruler to draw the tangent line straight.)

3. Repeat steps 1 and 2 for all of the remaining greatest elongations. (Accuracy is very important to the final result.)

4. When all tangent lines have been drawn very carefully, sketch in the orbit, making certain that it touches each tangent line at one point only.  Draw as smooth a curve as possible to represent Mercury's orbit.

5. Draw a line across the orbit representing the major axis.  Remember, the major axis must divide the orbit into two equal parts, it must go across the widest part of the orbit, it must pass through the sun and through the center of the orbit, and it must touch the orbit at the orbit's closest point to the sun (perihelion) and at its farthest point from the sun (aphelion).

6. Draw a line across the orbit representing the minor axis.  Remember, the minor axis must be perpendicular to the major axis and must pass through the center of the orbit.



Questions (To be answered on the back of the drawing)

l. Calculate the eccentricity of Mercury's orbit.  (The measured distance from the sun to the center of the orbit divided by the length of the semi-major axis.)

2. Describe what the eccentricity tells you about Mercury's orbit.

3. Measure and record the lengths of the major and minor axes of mercury's orbit. Which axis is longer?

4. What does the measurement and answer to question 3 tell you about Mercury's orbit?



      Poor old Tycho Brahe has left you 19 observations of Mercury when it was at greatest elongation (that is, at its greatest angle from the sun as seen from the earth).  From Nick Copernicus' publication on the orbits of the planets you know that you are looking tangent to a planet's orbit when you observe it at greatest elongation.  All you need to do is draw these tangents in the appropriate directions at the appropriate places with respect to the earth's orbit.  Once you have the outer perimeter, only a deft hand and keen eye are required to draw in, the actual orbit.

      Remember:  When the planet is east of the sun (such as at greatest eastern elongation), it seems to be to the left of the sun in the sky, and when it is west of the sun, to the right.  The angle is always measured from the position of the earth in its orbit on the appropriate date.

Answer 3

Haha I did this my junior year of high school of Honors Physics, and a little last year in AP Physics.

The area of a planet's orbit is as follows

A=(θ /360 degrees)π r²

That would be read as follows

Area=(number of degrees the planet passes through...if it isn't doing a complete revolution/360 degrees) times (Pi) times (the average radius of the for the orbit being measured squared)

*The average radius of the orbit is not necessarily for one complete revolution.

Thanks for copying MY ENTIRE EQUATIONS aniket patil!!!vvv

Answer 4

No-one else seems to have given you the correct answer, so here goes.

The area of an ellipse is πab where a is the semi-major axis and b is the semi-minor axis. b² = a²(1-e²) where e = eccentricity. For Mercury, a = 0.307499 A.U. and e = 0.205630. Therefore b = 0.300928 A.U. The area of the orbit is π x 0.307499 x 0.300928 = 0.290707 A.U.²

1 A.U. = 149,597,870 km so 1 A.U.² = 2.238 x 10^16 km². The area of Mercury's orbit is therefore 6.506 x 10^15 km².

Answer 5

Calculate the average radius (being that the orbit is elliptical), then use Pi * r2 (Pi multiplied by average radius squared)

I would calculate the average radius by adding the shortest radius (latitude) and longest (longitude) and dividing by two.