# 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 it's current orbit to notice a temperature change?

Mar 7, 2016

If the Earth was a mile closer, temperature would increase by 5.37\times10^{-7} %.
For the change in temperature to be noticeable, Earth would have to be 0.7175% closer to the sun.

#### Explanation:

The average surface temperature of the Earth as a result of it bathed by sunlight can be calculated by assuming that both the Earth and the Sun are blackbodies and the Earth is in radiative equilibrium with the Sun.

Note:If you want to know how, ask this as a separate question.

${T}_{E} = \setminus \sqrt{{R}_{S} / \left(2 r\right)} {T}_{S}$ ...... (1)

${T}_{E}$ - Radiative equilibrium temperature of the Earth,
${T}_{S} = 5778 \setminus \quad K$ - Surface temperature of the Sun,
${R}_{S} = 4.32686 \setminus \times {10}^{5} \setminus \quad m i$ - Radius of the Sun,
$r = 9.296 \setminus \times {10}^{7} \setminus \quad m i$ - Earth-Sun distance.

Substituting the numerical values for ${T}_{S} , {R}_{S}$ and $r$ we get
${T}_{E} = 278.74 \setminus \quad K = + {5.59}^{o} C$

We are interested in the following questions -
(1) How would ${T}_{E}$ change for a given change in $r$?
(2) What should be the change in $r$ to accomplish a given change in ${T}_{E}$.

To answer these questions, holding ${R}_{s}$ and ${T}_{s}$ as constants find the rate at which ${T}_{E}$ changes with $r$. For this rearrange equation 1 as follows,

${T}_{E} \setminus \sqrt{r} = \setminus \sqrt{{R}_{s}} . {T}_{S} =$ constant.
Differentiating once,

T_E/(2\sqrt{r})dr+\sqrt{r}dT_E=0; \qquad => (dT_E)/T_E=-1/2 (dr)/r

The change in Earth's surface temperature $\setminus \Delta {T}_{E}$ and the Earth-Sun distance $\setminus \Delta r$ are so tiny that we can write this differential as difference equation,

$\frac{\setminus \Delta {T}_{E}}{T} _ E = - \frac{1}{2} \frac{\setminus \Delta r}{r}$

Question 1: \Delta r = 1.0 \quad mi; \qquad r = 9.296\times10^7\quad mi
$\frac{\setminus \Delta {T}_{E}}{T} _ E = - \frac{1}{2} \frac{\setminus \Delta r}{r} = - 5.37 \setminus \times {10}^{- 9}$
Change in Temperature (in percentage ) is : (\DeltaT_E)/T_E\times100%
When the distance decrease by a mile the temperature on Earth increases by 5.37\times10^{-7} %.

Question 2: Assuming a change of $1 K$ as "noticeable", \DeltaT_E=1^o K; qquad T_E=278.74^oK
$\frac{\setminus \Delta r}{r} = - 2 \frac{\setminus \Delta {T}_{E}}{T} _ E = 7.175 \setminus \times {10}^{- 3}$
Change in distance (in percentage) is : ((\Delta r)/r)\times 100%
When the Earth-Sun decreases by 0.7175%, the surface temperature of Earth would increase by $1 K$

Mar 7, 2016