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?

Answer 1

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.

By assuming that the Earth and the Sun are both blackbodies and that the Earth is in radiative equilibrium with the Sun, the average surface temperature of the Earth as a result of its exposure to sunlight can be computed.

Note: Please ask this as a separate question if you would like to know how.

#T_E=\sqrt{R_S/(2r)}T_S# ...... (1)
#T_E# - Radiative equilibrium temperature of the Earth, #T_S = 5778\quad K# - Surface temperature of the Sun, #R_S = 4.32686\times10^5\quad mi# - Radius of the Sun, #r = 9.296\times10^7 \quad mi# - Earth-Sun distance.
Substituting the numerical values for #T_S, R_S# and #r# we get #T_E = 278.74\quad K = +5.59 ^oC#
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\sqrt{r} = \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 #\DeltaT_E# and the Earth-Sun distance #\Delta r# are so tiny that we can write this differential as difference equation,
#(\DeltaT_E)/T_E=-1/2 (\Deltar)/r#
Question 1: #\Delta r = 1.0 \quad mi; \qquad r = 9.296\times10^7\quad mi# #(\DeltaT_E)/T_E = -1/2(\Delta r)/r = -5.37\times10^{-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# #(\Delta r)/r = -2(\DeltaT_E)/T_E= 7.175\times10^{-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#
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Answer 2

A simpler answer. almost anything.

The Sun is actually closer in winter and farther away in summer (for the Northern hemisphere) due to the annual 5 million km variation in the Sun-Earth distance. However, this effect has little effect on the seasons. The year might actually be a few milliseconds shorter.

Because orbits are chaotic, the cumulative effects would have unpredictable effects in the Earth's orbit at very long times (billions of years).

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Answer 3

The Earth would experience a small increase in temperature if it were one mile closer to the Sun, but this would probably be insignificant. Several million miles would be required for the Earth to experience a noticeable change in temperature.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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