# Could I make a planet with these parameters?

I want a planet that is twice the diameter of earth's but has very similar gravity, atmosphere, and day length. Would doubling the rotational speed of the planet meet these parameters without notable down sides?

# Relation

TL;DR:

• Double the radius = Double the rotational speed.
• Double the radius = Half the density.

## Rotation

The relation between the diameter of a sphere and its circumference is linear:

$$c = d \times \pi$$

If you double the diameter of Earth, its circumference will also double. So in order to archive the same day time duration, you will need also to double the rotational speed.

## Gravity

But gravity is different:

$$g = \frac{G \times M}{r^2}$$

Where:

• $G$ is the gravitational constant. $G = 6.674 \times 10^{-11} \text{ N} \times \text{kg}{^-2} \times m^2$
• $M$ is the mass of the Earth.
• $r$ is the radius (half diameter) of Earth.

If you double the diameter, it goes through the square ($r^2$), and reduce the gravity 4 times.

So you must increase the mass of the planet 4 times in order to archive the same gravity.

But remember that mass is: $$M = V \times p$$ Where:

• $p$ is density: $p = \frac{M}{V}$
• $V$ is volume: (in sphere) $V=\frac{4}{3}\times\pi\times r^3$

So, if you double the diameter, it goes through the cube ($r^3$), and increase the mass 8 times, not 4 (like you need). In other words, If you double the diameter, you must half the density.

If you can't understand that you maybe should see that gravity can also be expressed as: $g = \frac{4\pi}{3}\times G\times p\times r$
$g = \frac{4\pi}{3}\times G\times\frac{M}{V}\times r$
$g = \frac{4\pi}{3}\times G\times\frac{M}{\frac{4}{3}\times\pi\times r^3}\times r$
$g = \frac{G\times M}{r^2}$

# Effects

This are some effects that this will produce:

• Due to the decrease of density in order to maintain the same gravity, your planet must be done almost completly of water and lighter elements. Thanks @Alexander for pointing that, you can check that in this link he post: 2.76 g/cm3 will be the density of the planet. (2.6 g/cm3 for an water like terrestial planet).
• The Earth will increase its eccentricity. That means it will have a less spherical shape because the middle of the Earth will be a bit wider. It will become a bit oblate spheroid.
• The precession of the Earth’s axis would change. The oblate spheroid shape will provoke bigger gravitational differences of the Sun on Earth, thas would make the axis precess even faster.
• Is possible that the axial tilt of the Earth would increase. This would mean that winters would be colder and summers would be hotter.
• This will increase the centrifugal force which means we would weight less in the equator because this acceleration counteracts a bit the gravity (1%?). The kinetic energy of the equator will increase quadratically.
• That will move a lot of water from the poles to the equator line raising the sea level. The equator zone will be drowned in water while the near polar zones will run out of water.
• Due to the faster spin, the decrease of gravity and rise of water at the equator, we could expect an increase of heavy moisture, dense fog, heavy clouds and constant rains.
• The increase of sea level and the decrease of equatorial gravity could produce much higher tides.
• The Earth's surface will rotate faster than its atmosphere so stronger winds will appear. That translate into stronger extreme weathers being more destructive, like hurricanes which will spin faster. Again this effect is produced because the equator line spins faster than the poles, provoking the Coriolis force. And the increase of rotation speed will always increase the coriolis effect.

Also note that:

• The increase of rotational speed isn't enough to produce massive earthquakes due to increase speed of tectonic plates.

Source:

• You didn't put in numbers, but the resulting density of this world will be 2.76 g/cm3, which is quite low for an Earth type planet. – Alexander Jul 20 '18 at 21:26
• @Alexander, maybe increase the water ratio of the planet and decrease heavy metals like uranium, lead or iron? – Ender Look Jul 21 '18 at 0:09
• Yes, but Planetary calculator that I used suggests that planet would be almost entirely made of water. – Alexander Jul 21 '18 at 0:35
• @Alexander impresive – Ender Look Jul 21 '18 at 0:37
• It should be clarified that “rotational speed” (i.e meters per second) is not angular speed (I.e radians per second). The planets angular speed still must be ~24 hours per revolution to match Earths day. – cms Jul 23 '18 at 3:45

Yes, it's possible. You'd have to change the density.

$$g = \frac{Gm}{r^2}$$

Where

• $g$ is Earth's surface gravity.
• $G$ is the gravitational constant.
• $m$ is the mass of the planet.
• $r$ is the radius.
• $g = 9.8 \text{ m/s}^2$
• $G = 6.674×10^{−11}\text{ N kg}^{–2}\text{ m}^2$

Earth’s radius is 6,378 km, which means the radius of this super-earth is 12,756 km or r = 12,756,000 meters

Solve for m.