- #1
alexk307
- 27
- 0
how much mass would it take to create its own gravity that's substantial enough to hold a human on it. And what is relationship between amount of mass and the amount of gravity it exerts?
Yes and no. Ignoring relativity, if the Earth had a spherical mass distribution, F=GmM/r^2 would be exact for any point on or above the Earth's surface. This is a direct consequence of Newton's Shell Theorem.gonegahgah said:F=GmM/r^2 is an approximation that works better the further away you are.
alexk307 said:how much mass would it take to create its own gravity that's substantial enough to hold a human on it. And what is relationship between amount of mass and the amount of gravity it exerts?
gonegahgah said:What this means is that the gravity we feel on the Earth is not linearly proportional to its mass. A denser planet will be smaller and exert a higher proportion of gravity per radius than will a less dense planet.
uart said:Hi Alex. You really need to think about exactly what you mean by the term "enough to hold a human on it". In the absence of any other nearby bodys any non-zero attractive force (which is of course what gravity is) will be enough to "hold a human".
QuantumPion said:My reasoning is that if the escape velocity is high enough so that you can't launch yourself into orbit with just your legs, then that object can hold a person. So what is the average jumping force of a person?
What this means is that the gravity we feel on the Earth is not linearly proportional to its mass. A denser planet will be smaller and exert a higher proportion of gravity per radius than will a less dense planet
QuantumPion said:The density of a planet does not affect its gravitational pull for a given mass. The greater density means the planet must have a smaller radius, which means the force of gravity at the surface will be stronger because the surface is closer to the center of mass. At a fixed distance from the center of mass, you will experience the same gravitational pull regardless of density.
uart said:Ok that gives a more specific definition of "hold a human".
Lets assume were talking about a roughly spherical lump of rock like an asteriod with a similar density to the Earths mantle, (about 3.5 E3 kg/m^2). Let's also assume that a person can jump to a height (on earth) of about one meter, corresponding to a velocity of about sqrt(2g). That is, v^2 approx equal to 20 (using all SI units, meter, kg, sec).
The escape velocity from our "rock" is v^2 = 2GM/r so in order to "hold our human" we require
2GM/r > 20 or GM/r > 10
Using M=4/3 pi r^3 rho we get r^2 > 30/(4 pi rho G).
Using numerical values of G=6.67E-11 and rho=3.5E+3 we get the minimum radius of,
r = 3200 meters
and the corresponding mass is
M = 4.8 E+14 kg
BTW That's just to avoid escape from jumping vertically. You'd still have to be careful not to run and jump because you could go into low Earth orbit (I mean low rock orbit) at only 71% of the escape velocity.
:Edit: The above is wrong because I underestimated how fast you could launch yourself vertically in a low gravity situation. Obviuosly the "typical" value of sqrt(20) that I calculated for Earth gravity is going to be a big under-estimate for the low g situation.
Say you are just limited by the speed of your leg kick, you might launch at more like 10 m/s. Taking this value and re-doing the above calcs gives GM/r > 50 and the numerical values of :
r = 7150 meters
M = 5.4 E+15 kg
Loren Booda said:gonegahgah
But isn't "A [uniformly] denser planet will be smaller and exert a higher proportion of gravity per radius than will a [uniformly] less dense planet" true? Gravity per radius and density are both proportional to 1/r3.
gonegahgah said:And this same thing flows through to each point on the planet. You can (hopefully) clearly see from this that the greater density planet will exert a greater attraction from the same distance from its centre than will the less dense planet despite them having the same mass.
The other thing you would have to acknowledge is that if we were able to squeeze the mass of the Earth into a column as long as the Earth's diametre and only 30cm across and stood on one end that we would feel greater weight than we do with the Earth in its spherical shape. This is because more of the mass would be directly below us; rather than off to the sides partly cancelling itself out.
http://en.wikipedia.org/wiki/MoonThe Moon's diameter is 3,474 km,[6] a little more than a quarter that of the Earth. This means that the Moon's volume is about 2 percent that of Earth and the pull of gravity at its surface about 17 percent that of the Earth.
QuantumPion said:My reasoning is that if the escape velocity is high enough so that you can't launch yourself into orbit with just your legs, then that object can hold a person. So what is the average jumping force of a person?
The amount of mass an object has is directly proportional to the strength of its gravitational pull. This means that the more mass an object has, the stronger its gravitational force will be.
Any amount of mass, no matter how small, will create some level of gravity. However, for the gravity to be strong enough to have a noticeable effect, the object would need to have a significant amount of mass, such as a planet or star.
No, mass is a necessary component for an object to have gravity. Objects with no mass, such as photons, do not have gravitational pull.
While size and mass are related, the strength of an object's gravity is determined by its mass, not its size. For example, a small but dense object, like a black hole, can have a much stronger gravitational pull than a larger but less dense object, like a planet.
Yes, the amount of gravitational force an object exerts can be increased by adding more mass to it. This can be seen with the Earth's moon, which has less mass than the Earth, and therefore has less gravitational force.