If we had an earth like planet except it had a 50 - 100 percent bigger radius than our earth what are some ways that it could have have similar gravity when compared to our own planet?
P.S. Atmospheric pressure is the same as earth
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1$\begingroup$ Half the density would be a good start $\endgroup$ – nzaman Mar 29 '18 at 16:30
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2$\begingroup$ What does 50% or 100% bigger mean? 1.5 to 2 times the volume of Earth, 1.5 to 2 times the surface area of Earth, 1.5 to 2 times the mass of Earth, or 1.5 to 2 times the radius of Earth? $\endgroup$ – AlexP Mar 29 '18 at 16:30
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$\begingroup$ Simply change the composition of the planet, use heavier or lighter liquids and metals... for example the earth is mostly iron $\endgroup$ – Ekaen Mar 29 '18 at 16:30
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$\begingroup$ AlexP I think i fixed the question for you $\endgroup$ – Christopher Void Mar 29 '18 at 16:32
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6$\begingroup$ Possible duplicate of What's the biggest reasonable natural planet or moon with Earth-like surface gravity? $\endgroup$ – The Square-Cube Law Mar 29 '18 at 16:51
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If you increase the radius, you actually have to increase the mass of the planet in order to feel the same gravity at the surface. Newton's equation for gravity is dependent on the radius between the center of two masses (say, the planet and a person standing on it):
$$ F_g = \frac{G m_1 m_2}{r^2} $$
If we want the force of gravity ($F_g$) to be the same for a planet with double the radius of Earth, we get an equation that looks like this:
$$ \frac{G m_1 m_2}{r^2} = \frac{G m_1 m_3}{(2r)^2} $$
where $m_1$ is the mass of our object on the planet's surface, $m_2$ is the mass of Earth, $m_3$ is the unknown mass of our new planet, $r$ is the radius of Earth, and $G$ is our gravitational constant. We can divide $G$ and $m_1$ from both sides to get
$$ \frac{m_2}{r^2} = \frac{m_3}{(2r)^2} $$
$$ m_3 = m_2 (\frac{2r}{r})^2 $$
$$ m_3 = m_2 2^2 $$
So the mass of planet with twice the radius of Earth would have to be four times the mass of Earth to have the same gravity at the surface. A planet with a 50% larger radius would have to have 2.25 times the mass. However, if you were to double the planet's radius, you would increase its volume 8 times while increasing its mass 4 times so you would have to halve its density. This might be achieved by replacing a significant amount of the iron in the Earth's mantle with magnesium and aluminum (maybe, I'm not a geologist).
The atmospheric pressure would be the same under these conditions, assuming that the height of the atmosphere were the same. But that being said, having larger characteristic length scales and curvature would change the way that weather occurs on this planet, which is to say I imagine it would be more boring. Probably not too much, though.
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If the planet had twice the radii of the Earth (2 x 6378 km), the planet would have to have 4 times the mass in order to have the same acceleration due to gravity on the surface.
The formula for acceleration due to gravity is:
$$g = \frac{G * M}{R^2}$$
Where:
g is the acceleration due to gravity.
G is the Universal Gravitational Constant (G)
M is the mass of the object (e.g. planet)
R is the distance to the center of mass of the object
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Gravity is about mass not size, so if the planet was made of less dense materials it would have less gravity in relation to it's size.
Easiest way would probably be to replace some of it's iron core with something less dense, or just give it a smaller iron core. (I'm assuming you want an iron core so that it will have a magnetic field to shield it from being sun fry)
Atmospheric pressure is not directly related to size or gravity either, Venus is comparable in size to Earth but has a much greater atmospheric pressure by many magnitudes.
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1$\begingroup$ The formula for gravitational force includes mass and the square of the distance. The size of a planet determines the surface's distance from the center, so size does matter. $\endgroup$ – Kyle A Apr 4 '18 at 4:51
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$\begingroup$ @KyleA good point, I was trying to keep it very simple since the OP doesn't seem to know a lot about it. $\endgroup$ – Kilisi Apr 4 '18 at 7:11
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$\begingroup$ Atmospheric pressure does have a direct relationship with gravity. First, pressure is force divided by area. In the case of atmospheric pressure, the force is gravitational. Greater gravity = greater pressure. Also, pressure with height changes according to the hydrostatic balance (dP/dz = density*g). So a higher gravity planet would have higher surface pressure and a shallower curve of pressure decrease with altitude. $\endgroup$ – Ryan Clare Apr 20 '18 at 3:29
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$\begingroup$ @RyanClare I'll bow to your more indepth knowledge. $\endgroup$ – Kilisi Apr 23 '18 at 3:15
