###Notes### The source of a number of the values I've used for Earth comparisons is the NASA Planetary Factsheet for Earth. For $g$ in particular is has this definition:

If we normalise $g$ so that it is instead $10\text{ } m/s^2$, to make these values easier to parse, then our required density becomes $35,778.07\text{ } g/cm^3$ and our comparison becomes:

For example, right at the surface $g = 10\text{ m/s}^{2}$, but, only 2m out from the planet's surface it changes to $g = 9.960\text{ m/s}^{2}$, and, were we to have a 2-3 story building, approximately 10m high, $g = 9.803\text{ m/s}^{2}$.

Notes

The source of a number of the values I've used for Earth comparisons is the NASA Planetary Factsheet for Earth. For $g$ in particular is has this definition:

There are a few issues with having "planets" this small

1. When your planet has a radius of 1000 m, the height of an adult human, is a noticeable percentage of the radius (on the order of 0.15 - 0.2%) and small buildings are closer to 1% of the radius!

To work out the surface gravity at the surface, we need to use the following equation

$$g = \frac{GM}{R^2}$$

The units of $g$ are $\text{m/s}^{2}$.

The components of this equation are:

• $G = 6.67259\text{ }*\text{ }10^{-11} \text{ ; units: }m^3\text{ }kg^{-1}\text{ }s^{-2}$
• $R = 1000 \text{ ; units: }m$
• $M = \rho\text{ } * \text{ Volume}\text{ ; units: }kg$
  • $\text{Volume} = \frac{4}{3} \pi R^3\text{ ; units: } m^3$

Substituting this all in we get:

$$g = \frac{6.67259\text{ }*\text{ }10^{-11}\text{ }*\text{ }\rho\text{ } * \frac{4}{3} \pi R^3}{R^2}\text{ ; units: } m/s^2 $$ $$ = 6.67259\text{ }*\text{ }10^{-11}\text{ }*\text{ }\rho\text{ }* \frac{4}{3} \pi R\text{ ; units: } m/s^2 $$ $$ = 6.67259\text{ }*\text{ }10^{-8}\text{ }*\text{ }\rho\text{ }* \frac{4}{3} \pi \text{ ; units: } m/s^2$$

So the key variable for targeting a particular gravity is $\rho$. If we want to target a $g$ close to that of Earth ($9.798\text{ }m/s^{2}$^{source: NASA factsheet}), then we need a value of $\rho = 35,055 \text{ }g/cm^3$ (ie approximately the density of some black holes and white dwarf stars!). It also gives our planet a mass of $1.47 * 10^{17}$ kg (when we change the radius later for analysing surface gravity changes, we will need to keep the planetary mass constant).

If we go with that, then we run into a separate issue...that the force of gravity changes appreciably over scales as small as the human body (which would be an issue for small things like distribution of blood over the body).

For example, right at the surface $g = 9.798\text{ m/s}^{2}$, but, only 2m out from the planet's surface it changes to $g = 9.759\text{ m/s}^{2}$, and, were we to have a 2-3 story building, approximately 10m high, $g = 9.605\text{ m/s}^{2}$.

2. If we make the density of the planet low, to counteract this drastic change in surface gravity over different parts of the human body, we will make the escape velocity of the "planet" significantly lower

The equation to calculate escape velocity is:

$$v_\text{escape} = \sqrt{\frac{2GM}{R}}\text{ ; units: } m/s$$

If we were to change the density of our small planet, down to that of Earth ($5.51\text{ } g/cm^{3}$^{source: NASA factsheet}), then our mass becomes $2.3 * 10^{13}$ kg and we get $g = 0.00154\text{ m/s}^{2}$.

From the perspective of our corporation, this is much more desirable, as they need to source a significantly smaller mass of material (by a factor of 10,000!).

However, if we have a surface gravity that low, then working through the numbers we end up with $v_\text{escape} = 1.755\text{ } m/s = 6.32\text{ } km/h$. This is low enough that a human would likely easily be able to reach that speed. Usain Bolt has achieved speeds of $10.44\text{ m/s}$ or $37.58\text{ km/h}$, so a speed of $6.32\text{ } km/h$ is certainly within the capability of a regular human.

Conclusion

The primary parameters that we would need to balance are the radius and the density of the planet. To mitigate the most severe gravitational and escape velocity problems, we would need our planet to be significantly larger and have a surface gravity pretty significantly lower than that of Earth.

There are a few issues with having "planets" this small

1. When your planet has a radius of 1000 m, the height of an adult human, is a noticeable percentage of the radius (on the order of 0.15 - 0.2%) and small buildings are closer to 1% of the radius!

To work out the surface gravity at the surface, we need to use the following equation

$$g = \frac{GM}{R^2}$$

The units of $g$ are $\text{m/s}^{2}$.

The components of this equation are:

• $G = 6.67259\text{ }*\text{ }10^{-11} \text{ ; units: }m^3\text{ }kg^{-1}\text{ }s^{-2}$
• $R = 1000 \text{ ; units: }m$
• $M = \rho\text{ } * \text{ Volume}\text{ ; units: }kg$
  • $\text{Volume} = \frac{4}{3} \pi R^3\text{ ; units: } m^3$

Substituting this all in we get:

$$g = \frac{6.67259\text{ }*\text{ }10^{-11}\text{ }*\text{ }\rho\text{ } * \frac{4}{3} \pi R^3}{R^2}\text{ ; units: } m/s^2 $$ $$ = 6.67259\text{ }*\text{ }10^{-11}\text{ }*\text{ }\rho\text{ }* \frac{4}{3} \pi R\text{ ; units: } m/s^2 $$ $$ = 6.67259\text{ }*\text{ }10^{-8}\text{ }*\text{ }\rho\text{ }* \frac{4}{3} \pi \text{ ; units: } m/s^2$$

So the key variable for targeting a particular gravity is $\rho$. If we want to target a $g$ close to that of Earth ($9.798\text{ }m/s^{2}$^{source: NASA factsheet}), then we need a value of $\rho = 35,055 \text{ }g/cm^3$ (ie approximately the density of some black holes and white dwarf stars!). It also gives our planet a mass of $1.47 * 10^{17}$ kg (when we change the radius later for analysing surface gravity changes, we will need to keep the planetary mass constant).

If we go with that, then we run into a separate issue...that the force of gravity changes appreciably over scales as small as the human body (which would be an issue for small things like distribution of blood over the body).

For example, right at the surface $g = 9.798\text{ m/s}^{2}$, but, only 2m out from the planet's surface it changes to $g = 9.759\text{ m/s}^{2}$, and, were we to have a 2-3 story building, approximately 10m high, $g = 9.605\text{ m/s}^{2}$.

2. If we make the density of the planet low, to counteract this drastic change in surface gravity over different parts of the human body, we will make the escape velocity of the "planet" significantly lower

The equation to calculate escape velocity is:

$$v_\text{escape} = \sqrt{\frac{2GM}{R}}\text{ ; units: } m/s$$

If we were to change the density of our small planet, down to that of Earth ($5.51\text{ } g/cm^{3}$^{source: NASA factsheet}), then our mass becomes $2.3 * 10^{13}$ kg and we get $g = 0.00154\text{ m/s}^{2}$.

From the perspective of our corporation, this is much more desirable, as they need to source a significantly smaller mass of material (by a factor of 10,000!).

However, if we have a surface gravity that low, then working through the numbers we end up with $v_\text{escape} = 1.755\text{ } m/s = 6.32\text{ } km/h$. This is low enough that a human would likely easily be able to reach that speed. Usain Bolt has achieved speeds of $10.44\text{ m/s}$ or $37.58\text{ km/h}$, so a speed of $6.32\text{ } km/h$ is certainly within the capability of a regular human.

Conclusion

The primary parameters that we would need to balance are the radius and the density of the planet. To mitigate the most severe gravitational and escape velocity problems, we would need our planet to be significantly larger and have a surface gravity pretty significantly lower than that of Earth.

###Notes### The source of a number of the values I've used for Earth comparisons is the NASA Planetary Factsheet for Earth. For $g$ in particular is has this definition:

Equatorial gravitational acceleration at the surface of the body or the 1 bar level, not including the effects of rotation, in meters/(second^2)

Defined here.

There are a few issues with having "planets" this small

1. When your planet has a radius of 1000 m, the height of an adult human, is a noticeable percentage of the radius (on the order of 0.15 - 0.2%) and small buildings are closer to 1% of the radius!

To work out the surface gravity at the surface, we need to use the following equation

$$g = \frac{GM}{R^2}$$

The units of $g$ are $\text{m/s}^{2}$.

The components of this equation are:

• $G = 6.67259\text{ }*\text{ }10^{-11} \text{ ; units: }m^3\text{ }kg^{-1}\text{ }s^{-2}$
• $R = 1000 \text{ ; units: }m$
• $M = \rho\text{ } * \text{ Volume}\text{ ; units: }kg$
  • $\text{Volume} = \frac{4}{3} \pi R^3\text{ ; units: } m^3$

Substituting this all in we get:

$$g = \frac{6.67259\text{ }*\text{ }10^{-11}\text{ }*\text{ }\rho\text{ } * \frac{4}{3} \pi R^3}{R^2}\text{ ; units: } m/s^2 $$ $$ = 6.67259\text{ }*\text{ }10^{-11}\text{ }*\text{ }\rho\text{ }* \frac{4}{3} \pi R\text{ ; units: } m/s^2 $$ $$ = 6.67259\text{ }*\text{ }10^{-8}\text{ }*\text{ }\rho\text{ }* \frac{4}{3} \pi \text{ ; units: } m/s^2$$

So the key variable for targeting a particular gravity is $\rho$. If we want to target a $g$ close to that of Earth ($9.798\text{ }m/s^{2}$), then we need a value of $\rho = 35,055 \text{ }g/cm^3$ (ie approximately the density of some black holes and white dwarf stars!). It also gives our planet a mass of $1.47 * 10^{17}$ kg (when we change the radius later for analysing surface gravity changes, we will need to keep the planetary mass constant).

If we go with that, then we run into a separate issue...that the force of gravity changes appreciably over scales as small as the human body (which would be an issue for small things like distribution of blood over the body).

For example, right at the surface $g = 9.798\text{ m/s}^{2}$, but, only 2m out from the planet's surface it changes to $g = 9.759\text{ m/s}^{2}$, and, were we to have a 2-3 story building, approximately 10m high, $g = 9.605\text{ m/s}^{2}$.

2. If we make the density of the planet low, to counteract this drastic change in surface gravity over different parts of the human body, we will make the escape velocity of the "planet" significantly lower

The equation to calculate escape velocity is:

$$v_\text{escape} = \sqrt{\frac{2GM}{R}}\text{ ; units: } m/s$$

If we were to change the density of our small planet, down to that of Earth ($5.51\text{ } g/cm^{3}$), then our mass becomes $2.3 * 10^{13}$ kg and we get $g = 0.00154\text{ m/s}^{2}$.

From the perspective of our corporation, this is much more desirable, as they need to source a significantly smaller mass of material (by a factor of 10,000!).

However, if we have a surface gravity that low, then working through the numbers we end up with $v_\text{escape} = 1.755\text{ } m/s = 6.32\text{ } km/h$. This is low enough that a human would likely easily be able to reach that speed. Usain Bolt has achieved speeds of $10.44\text{ m/s}$ or $37.58\text{ km/h}$, so a speed of $6.32\text{ } km/h$ is certainly within the capability of a regular human.

Conclusion

The primary parameters that we would need to balance are the radius and the density of the planet. To mitigate the most severe gravitational and escape velocity problems, we would need our planet to be significantly larger and have a surface gravity pretty significantly lower than that of Earth.

There are a few issues with having "planets" this small

1. When your planet has a radius of 1000 m, the height of an adult human, is a noticeable percentage of the radius (on the order of 0.15 - 0.2%) and small buildings are closer to 1% of the radius!

To work out the surface gravity at the surface, we need to use the following equation

$$g = \frac{GM}{R^2}$$

The units of $g$ are $\text{m/s}^{2}$.

The components of this equation are:

• $G = 6.67259\text{ }*\text{ }10^{-11} \text{ ; units: }m^3\text{ }kg^{-1}\text{ }s^{-2}$
• $R = 1000 \text{ ; units: }m$
• $M = \rho\text{ } * \text{ Volume}\text{ ; units: }kg$
  • $\text{Volume} = \frac{4}{3} \pi R^3\text{ ; units: } m^3$

Substituting this all in we get:

$$g = \frac{6.67259\text{ }*\text{ }10^{-11}\text{ }*\text{ }\rho\text{ } * \frac{4}{3} \pi R^3}{R^2}\text{ ; units: } m/s^2 $$ $$ = 6.67259\text{ }*\text{ }10^{-11}\text{ }*\text{ }\rho\text{ }* \frac{4}{3} \pi R\text{ ; units: } m/s^2 $$ $$ = 6.67259\text{ }*\text{ }10^{-8}\text{ }*\text{ }\rho\text{ }* \frac{4}{3} \pi \text{ ; units: } m/s^2$$

So the key variable for targeting a particular gravity is $\rho$. If we want to target a $g$ close to that of Earth ($9.798\text{ }m/s^{2}$^{source: NASA factsheet}), then we need a value of $\rho = 35,055 \text{ }g/cm^3$ (ie approximately the density of some black holes and white dwarf stars!). It also gives our planet a mass of $1.47 * 10^{17}$ kg (when we change the radius later for analysing surface gravity changes, we will need to keep the planetary mass constant).

If we go with that, then we run into a separate issue...that the force of gravity changes appreciably over scales as small as the human body (which would be an issue for small things like distribution of blood over the body).

For example, right at the surface $g = 9.798\text{ m/s}^{2}$, but, only 2m out from the planet's surface it changes to $g = 9.759\text{ m/s}^{2}$, and, were we to have a 2-3 story building, approximately 10m high, $g = 9.605\text{ m/s}^{2}$.

2. If we make the density of the planet low, to counteract this drastic change in surface gravity over different parts of the human body, we will make the escape velocity of the "planet" significantly lower

The equation to calculate escape velocity is:

$$v_\text{escape} = \sqrt{\frac{2GM}{R}}\text{ ; units: } m/s$$

If we were to change the density of our small planet, down to that of Earth ($5.51\text{ } g/cm^{3}$^{source: NASA factsheet}), then our mass becomes $2.3 * 10^{13}$ kg and we get $g = 0.00154\text{ m/s}^{2}$.

From the perspective of our corporation, this is much more desirable, as they need to source a significantly smaller mass of material (by a factor of 10,000!).

However, if we have a surface gravity that low, then working through the numbers we end up with $v_\text{escape} = 1.755\text{ } m/s = 6.32\text{ } km/h$. This is low enough that a human would likely easily be able to reach that speed. Usain Bolt has achieved speeds of $10.44\text{ m/s}$ or $37.58\text{ km/h}$, so a speed of $6.32\text{ } km/h$ is certainly within the capability of a regular human.

Conclusion

The primary parameters that we would need to balance are the radius and the density of the planet. To mitigate the most severe gravitational and escape velocity problems, we would need our planet to be significantly larger and have a surface gravity pretty significantly lower than that of Earth.