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
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)?
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