How fast (in `m^(2) s^(-1))` is area swept out by (a) the radius from sun to earth ? (b) the radius from the sun to earth `= 1.496 xx 10^(11) m` , Distance of earth to moon `= 3.845 xx 10^(8) m` and period of revolution of moon `= 27 (1)/(3)` days.

Explanation

(a) Radius of earth orbit around sun
`R = 1.496 xx 10^(11) m` ,
Time period of revolution of earth around sun
`T = 365 days = 365 xx 24 xx 60 xx 60 s`
The rate at which the area is swept out by the radius from the sun to earth,
`(dA)/(dt) = (pi R^(2))/(T) = ((22//7)(1.496 xx 10^(11))^(2))/(365 xx 24 xx 60 xx 60)`
`= 2.23 xx 10^(15) m^(2) s^(-1)`
(b) Radius of moon orbit around earth,
`R = 3.845 xx 10^(8) m`
Time period of revolution of moon around earth
`T = 27(1)/(3) days = (82)/(3) xx 24 xx 60 xx 60 s`
The rate at which the area is swept out by radius from the earth to moon
`(dA)/(dt) = (pi R^(2))/(T) = ((22//7)(3.845 xx 10^(8))^(2))/((82//3) xx 24 xx 60 xx 60)`
`= 1.97 xx 10^(11) m^(2) s^(-1)`