The major axis is the diameter across the widest part. The semimajor axis is half that, and for a planet it's the average of the maximum and minimum distances from the Sun .
jkjnl
That can be calculated from Kepler's 3rd law which says if the period is T years the semimajor axis must be T2/3 astronomical units. So for a period of 12 years the s/m axis is 5.421 AU or 784 million km.
The eccentricity of that ellipse is 0.4 .
4 years
The sun is located at one of the focii of the ellipse that describes the orbit of a planet. The focus of an ellipse is not in the geometric center but on the major axis of the ellipse.
One of the parts of an ellipse is the length of its major axis. Half that is called the semimajor axis. Kepler's 3rd law says that the time to do one orbit is proportional to the 3/2 power of the semimajor axis. IF the semimajor axis is one astronomical unit the period is one year (the Earth). For a planet with a semimajor axis of 4 AUs the period would have to be 8 years, by Kepler-3.
One of the parts of an ellipse is the length of its major axis. Half that is called the semimajor axis. Kepler's 3rd law says that the time to do one orbit is proportional to the 3/2 power of the semimajor axis. IF the semimajor axis is one astronomical unit the period is one year (the Earth). For a planet with a semimajor axis of 4 AUs the period would have to be 8 years, by Kepler-3.
The major axis of an ellipse is its longest diameter, a line that runs through the center and both foci, its ends being at the widest points of the shape.The semi-major axis is one half of the major axis, and thus runs from the centre, through a focus, and to the edge of the ellipse. It represents a "long radius" of the ellipse, and is the "average" distance of an orbiting planet or moon from its parent body.
jkjnl
The major and minor axes of a circle are the same - either is any diameter. So a semimajor axis is half the diameter which is 12 cm.
Moment of inertia about x-axis for an ellipse is = pi*b^3*a /4. Where b is the distance from the center of the ellipse to the outside tip of the minor axis. a is the distance from the ceneter of the ellipse to the outside tip of the major axis. Moment of inertia about x-axis for an ellipse is = pi*b^3*a /4. Where b is the distance from the center of the ellipse to the outside tip of the minor axis. a is the distance from the ceneter of the ellipse to the outside tip of the major axis.
the period of revolution is related to the semimajor axis.... :)
yes
Yes.
A
The area of an ellipse with a major axis 20 m and a minor axis 10 m is: 157.1 m2
Ellipse formula, centered at the origin, where the vertical axis is the major axis: x2/b2 + y2/a2 = 1, a > b Since the major axis is 8, then a = 4. Since the minor axis is 4, then b = 2. Thus, the equation of the ellipse is: x2/4 + y2/16 = 1.