Asked by Mischa
Help! I have no idea how to do this.
The earth's orbit is nearly circular with a radius R=93x10^6 miles (the eccentricity is e=.017). Find the rate at which the earth's radial vector sweeps out area in units of ft^2/s. What is the magnitude of the vector J=rxr` for the earth (in units of squared feet per second)?
The earth's orbit is nearly circular with a radius R=93x10^6 miles (the eccentricity is e=.017). Find the rate at which the earth's radial vector sweeps out area in units of ft^2/s. What is the magnitude of the vector J=rxr` for the earth (in units of squared feet per second)?
Answers
Answered by
bobpursley
One knows the time it takes to go around the Sun.
Note that area swept out is PI*R^2/time
that is area in miles^2/year. Convert that to ft^2/sec
J= r x r'= R*dR/dTheta= R*R dTheta/dtime
= R^2 * 2PI/1year
convert that to ft^2/sec
Note that area swept out is PI*R^2/time
that is area in miles^2/year. Convert that to ft^2/sec
J= r x r'= R*dR/dTheta= R*R dTheta/dtime
= R^2 * 2PI/1year
convert that to ft^2/sec
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