# Shall the Moon be Bigger or Closer? [closed]

The last time I asked about the moon, I used a scenario where it has enlarged to a diameter of 2500 miles orbiting 200,000 miles from Earth.

But what if our moon has one difference or the other?

In one scenario, Earth's moon is the same size--2159 miles in diameter--but stands 200,000 miles from Earth.

In another, Earth's moon is the same distance from Earth--238,900 miles--but enlarged to 2500 miles in diameter.

Using either difference, would the nightscape look any different? How would tides and lunar cycles be affected?

## closed as unclear what you're asking by Thucydides, JDługosz, Youstay Igo, Brythan, GreenOct 4 '15 at 13:08

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

Let's first look at the apparent size. The apparent radius is proportional to the quotient of real radius and distance (as long as that quotient is sufficiently small, which it is for all your scenarios).

In your first case, the ratio is $2159/200\,000 \approx 0.0108$, while in the second case, the ratio is $2500/238\,900 \approx 0.0105$. So in the first scenario, the moon's diameter would be about 3% larger. That's less than the variations of the apparent diameter of our real moon (which are close to 13%), and I think most people are not even aware that the moon changes apparent size, so you can conclude that an even smaller difference would not be relevant. The same is true for the apparent brightness: For constant apparent diameter (i.e. radius $\propto$ distance), the fact that the larger moon reflects more sunlight (proportional to its radius, thus $\propto r^2$) is exactly cancelled out by the larger distance (the brightness is $\propto 1/d^2$), thus the apparent brightness is simply proportional to the apparent area of the moon, which again is quadratic in the radius. The bigger of your moons thus would be about 6% more bright, which certainly is not nothing, but in the end doesn't make a big difference.

Now about the gravitational effects: Assuming your moons have constant density, their mass goes with the third power of the radius. Since gravitation strength is $\propto d^2$, on first view this seems to imply that the tides would be much stronger for the larger, more distant moon. However the direct gravitation is cancelled out by the centrifugal force, and the tidal forces go $\propto 1/d^3$. Therefore the height of the tides is again coupled to the apparent diameter of the moon, however this time to the third power. Inserting your numbers gives that the tidal forces would be higher by about 10%. However note that the actual height of the tides is very much affected by the local situation (that's why tides are of different height at different places). Therefore I'm not sure how much those 10% higher tidal forces would affect the actual height of the tides.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}