Exercise 7. Cyclical changes in Earth’s solar orbit have been proposed as one ca

Question

Exercise 7.

Cyclical changes in Earth’s solar orbit have been proposed as one cause of the variations (cycles and fluctuations) in the planet’s average temperature. There are three separate effects, commonly known as Milankovitch cycles. The two most important are changes in the size of the tilt and the orientation of the tilt of Earth’s rotational (polar) axis, with respect to the direction perpendicular to the Earth’s orbital plane around the sun. These two effects require more advanced mathematics to study. However, the third effect can be studied in an idealized form: change in the mean distance between the earth and the sun, i.e. small changes in the size of 1 A.U. Assuming a circular orbit, by how much would the distance from Earth to the sun have to change in order to bring about a 1 K increase in To = 255 K?

Explanation / Answer

What we know in the question:

r0 (distance from the sun to Earth) = 1.50x10^11 m

T0 (black body temperature of Earth) = 255 K

T1=T0+1 = 256K

a (albedo of earth) = .3

Formula which will be used here:

Pa/(4*pi*r^2) = *T^4

Solution:

For the current distance r0 where To = 255K

Pa/(4*pi*r0^2) = *To^4

Put T0 = T to ease the calculation

Pa/(4*pi*r0^2) = *T^4

For the new distance, r1 with increased temperature T1

Pa/(4*pi*r1^2) = *T1^4

also we know that T1=T+1

thus:

Pa/(4*pi*r1^2) = *(1+T)^4

Therefore,

r1^2/ro^2 = Pa/(4 x pi x (T+1)^4 ) divided by Pa/(4*pi**T^4 )

r1^2/ro^2 = Pa/(4 x pi x (T+1)^4 ) multiplied by (4 x pi x T^4 )/Pa

P, a, 4, pi and are all constants

Cancelling top and bottom

r1^2/ro^2 = T^4 / (T+1)^4

r1^2 = ro^2 x T^4 / (T+1)^4

r1^2 = ro^2 x 255^4 / (255+1)^4

r1^2 = ro^2 x 255^4 / 256^4

r1^2 = ro^2 x 4.2282 x10^9 / 4.2949 x10^9

r1^2 = ro^2 x 0.9844

r1 = 1.5 x 10^11 x root [0.9844]

r1 =1.4882 x 10^11 m (a little closer)

we are looking for the difference (r0 - r1)

= 1.5 x 10^11 -1.4482 x 10^11

= 1.17 x 10^9 m

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