Question

If the earth were a mile closer to the sun how would it affect our temperature on earth? How far would the earth have to be from its current orbit to notice a temperature change?

Answer 1

Hint: The change in temperature of earth can be calculated with respect to the change in distance between earth and sun by considering earth and sun as black bodies and also considering that earth is in radiative equilibrium with the sun.

Complete step by step answer:
We know, the temperature of earth with respect to surface temperature of the sun can be calculated by the following formula:
${T_E} = \sqrt {\dfrac{{{R_S}}}{r}} {T_S}.....(1)$
Where,
${T_E}$, is the temperature of the earth which is in radiative equilibrium of the sun
${T_S}$, is the surface temperature if the sun
${R_S}$, is the radius of the sun
$r$, is the distance between the sun and the earth
Considering the values, ${T_S}$ and ${R_S}$ as constant, and differentiating ${T_E}$with respect to $r$ after rearranging the equation as:
${T_E}\sqrt r = \sqrt {{R_S}} {T_S}$
On differentiating, we obtain:
$\dfrac{{{T_E}}}{{2\sqrt r }}dr + \sqrt r {T_E} = 0$
$\Rightarrow \dfrac{{\Delta {T_E}}}{{{T_E}}} = - \dfrac{1}{2}\dfrac{{\Delta r}}{r}$
In the question, it is given that the distance is increased by one mile.
Therefore, we put: $r = 9.296 \times {10^7}$
Therefore, change in temperature is:
$\dfrac{{\Delta {T_E}}}{{{T_E}}} = - \dfrac{1}{2}\dfrac{{\Delta r}}{r} = - 5.37 \times {10^{ - 9}}$
Therefore, in percentage:
$\dfrac{{\Delta {T_E}}}{{{T_E}}} \times 100 = - 5.37 \times {10^{ - 7}}\%$
Therefore, with an increase in one-mile distance the temperature of the earth increases by$5.37 \times {10^{ - 7}}\%$.
We know, the value of $T$ is $5778K$, ${R_S} = 4.32386 \times {10^5}mi$, $r = 9.292 \times {10^7}mi$
These are constant values.
Putting these values in equation$(1)$, we get:
 ${T_E} = 278.74K$
Assuming the noticeable change in temperature being $1K$
Putting the values, in equation obtained after temperature is:
$\dfrac{{\Delta r}}{r} = - 2\dfrac{{\Delta {T_E}}}{{{T_E}}}$
On solving this, we obtain:
$\dfrac{{\Delta r}}{r} = 7.175 \times {10^{ - 3}}$
In terms of percentage, we get:
$\dfrac{{\Delta r}}{r} \times 100 = 0.7175\%$
Therefore, when the distance between the sun and earth decreases by $0.7175\%$, the surface temperature of the earth increases by $1K$.

Note: We know as Earth revolves around the sun in an elliptical orbit, the axis along which the earth rotates is slightly tilted as it spins on the top. Due to rotation, there occurs a change in angle through which it is tilted. Thus, the angle with which the sunlight hits the surface of the earth changes, this results in a change in seasons that we experience throughout the year.