Surface Gravity Versus Centrifugal Force

[Elen Sila]

So there's this asteroid called Sylvia. (It's primarily known for having two moons, Remus and Romulus, but that's beside the point.) Sylvia rotates about once every 18661 seconds, sidereally; and her mean radius is about 286 kilometers. That means that, at her equator, assuming approximate sphericity (obviously a big assumption, but that's also beside the point), she rotates at about 96.3 meters per second. But, assuming an object of negligible mass could revolve around her at a non-eccentric distance of exactly her mean radius (again, implausible, but let's not get hung up on that), that object would have an orbital period of (286000^3/(1.478*10^19)/(6.67428*10^-11))^(1/2)*pi*2≈30598 seconds, and an orbital velocity of about 58.7 meters per second.

 

If Sylvia's surface rotation speed is greater than her surface orbital velocity, shouldn't Sylvia have a negative surface gravity? If an astronaut wanted to land on her surface, wouldn't he have to use mountain climbing gear just to keep from flying off into space? Why does Wikipedia list Sylvia's surface gravity as 0.0729 meters per second per second, instead of accounting for centrifugal force in that figure? And does Wikipedia account for centrifugal force in its figure for the earth's gravity (9.780327 meters per second per second at sea level on the equator)? If so, why the inconsistency?

 

Anyway, my main question is this. I know how to calculate surface gravity on a non-rotating object. But how do you calculate Sylvia's (or any other body's) actual surface gravity, accounting for centrifugal force? By extension, how would you measure the artificial gravity in a rotating space station, in meters per second per second?

 

http://en.wikipedia...._%28asteroid%29

Pic related, it's Sylvia.

[Iggy]

Sylvia rotates about once every 18661 seconds, sidereally; and her mean radius is about 286 kilometers.

286 is a diameter. The small axis is the rotational axis. The equator should have about 192km radius which, you could check the math that I've haphazardly done, gives an orbital velocity of 72 m/s and a rotational velocity of 61 m/s.

[csmyth3025]

So there's this asteroid called Sylvia. (It's primarily known for having two moons, Remus and Romulus, but that's beside the point.) Sylvia rotates about once every 18661 seconds, sidereally; and her mean radius is about 286 kilometers. That means that, at her equator, assuming approximate sphericity (obviously a big assumption, but that's also beside the point), she rotates at about 96.3 meters per second. But, assuming an object of negligible mass could revolve around her at a non-eccentric distance of exactly her mean radius (again, implausible, but let's not get hung up on that), that object would have an orbital period of (286000^3/(1.478*10^19)/(6.67428*10^-11))^(1/2)*pi*2≈30598 seconds, and an orbital velocity of about 58.7 meters per second.

 

If Sylvia's surface rotation speed is greater than her surface orbital velocity, shouldn't Sylvia have a negative surface gravity? If an astronaut wanted to land on her surface, wouldn't he have to use mountain climbing gear just to keep from flying off into space? Why does Wikipedia list Sylvia's surface gravity as 0.0729 meters per second per second, instead of accounting for centrifugal force in that figure? And does Wikipedia account for centrifugal force in its figure for the earth's gravity (9.780327 meters per second per second at sea level on the equator)? If so, why the inconsistency?

 

Anyway, my main question is this. I know how to calculate surface gravity on a non-rotating object. But how do you calculate Sylvia's (or any other body's) actual surface gravity, accounting for centrifugal force? By extension, how would you measure the artificial gravity in a rotating space station, in meters per second per second?

The formula for calculation the centripetal (inward) force exerted on an orbiting body is:

 

 

Essentially the acceleration is v²/r.

 

The equatorial rotational velocity for 87 Sylvia is given as:

 

Sylvia is also a fairly fast rotator, turning about its axis every 5.18 hours (giving an equatorial rotation velocity of about 230 km/h or 145 mph). The short axis is the rotation axis.^[4] Direct images^[3] indicate that Sylvia's pole points towards ecliptic coordinates (β, λ) = (+62.6°, 72.4°) with only a 0.5° uncertainty, which gives it an axial tilt of around 29.1°. Sylvia's shape is strongly elongated.

(ref. http://en.wikipedia....characteristics )

 

In this case, the equatorial velocty would be (230,000 m/hr)/(3600) = ~63.89 m/s. The radius used for this calculation would be:

 

(63.89 m/s)/(2*pi/18,648 s) = 189,620 m = 189.62 km radius = ~380 km diameter. This is pretty close to the 384 km diameter long axis given in the Wikipedia article. The mass of 87 Sylvia is given as 1.478x10¹⁹ kg in this same article.

 

Using round numbers, the acceleration of gravity at ~190,000 meter radius =

 

(G*m)/r² = [(6.674x10^-11m³/kg*s²)(1.478x10¹⁹ kg)]/(1.9x10⁵ m)² = ~(9.86x10⁸ m³/s²)/(3.61x10¹⁰ m²) = ~0.027 m/s²

 

Since this number isn't even close to the 0.0729 m/s² equatorial surface gravity givenin the article (not even counting the effect of rotation), I've obviously done something wrong.

 

Help!

 

Chris

 

Edited to correct spelling errors

Edited September 3, 2011 by csmyth3025

[Elen Sila]

286 is a diameter.

 

... Wow. I'm so used to Wikipedia giving me the radii that I didn't even realize I was looking at the diameter.

I hate diameters so much. Radii are so much more useful and relevant. I keep 2pi saved to my calculator as a constant, and almost never use my calculator's Pi button.

 

The small axis is the rotational axis. The equator should have about 192km radius which, you could check the math that I've haphazardly done, gives an orbital velocity of 72 m/s and a rotational velocity of 61 m/s.

 

Yeah, I get 64.6 meters per second for the equatorial rotation speed, at an equatorial radius of 192 kilometers. The orbital period at that distance would be 16831 seconds, and the orbital velocity would be 71.7 meters per second. So I guess the rotation speed isn't greater than the orbital velocity, and no negative gravity would be present. I'd still like to know how to calculate centrifugal/centripetal force on a rotating gravitational body, though.

 

The formula for calculation the centripetal (inward) force exerted on an orbiting body is:

 

 

Essentially the acceleration is v²/r.

 

For the record, when you say "centripetal (inward) force", are you referring to, for example, the force of the floor of a rotating space station against the feet of an astronaut standing in it?

Also, when you say "orbiting body", do you mean "body in freefall" or "rotating body"?

 

Using round numbers, the acceleration of gravity at ~190,000 meter radius =

 

(G*m)/r² = [(6.674x10^-11m³/kg*s²)(1.478x10¹⁹ kg)]/(1.9x10⁵ m)² = ~(9.86x10⁸ m³/s²)/(3.61x10¹⁰ m²) = ~0.027 m/s²

 

Since this number isn't even close to the 0.0729 m/s² equatorial surface gravity givenin the article (not even counting the effect of rotation), I've obviously done something wrong.

 

Help!

 

(6.67384*10^-11)*(1.478*10^19)/192000^2, sure enough, is 0.026758 meters per second per second.

However, 192 kilometers is only the asteroid's widest equatorial radius. Its mean radius is 143 kilometers, and its polar radius is 116 kilometers.

If we plug in the mean radius, we get a surface gravity of 0.048237 meters per second per second; and if we plug in the polar radius, we get a surface gravity of 0.073305 meters per second per second.

If we reverse-figure from 0.0729 meters per second, keeping the stated mass, we get a radius of 116.322 kilometers.

 

I'm guessing the article is using the polar surface gravity; and since the polar surface gravity wouldn't be subject to the same centrifugal concerns, they wouldn't even have to factor that in. I'd still like to know how to calculate Sylvia's equatorial surface gravity though, accounting for centrifugal force.

Edited September 3, 2011 by Elen Sila

[csmyth3025]

...For the record, when you say "centripetal (inward) force", are you referring to, for example, the force of the floor of a rotating space station against the feet of an astronaut standing in it?

Also, when you say "orbiting body", do you mean "body in freefall" or "rotating body"?

 

 

 

(6.67384*10^-11)*(1.478*10^19)/192000^2, sure enough, is 0.026758 meters per second per second.

However, 192 kilometers is only the asteroid's widest equatorial radius. Its mean radius is 143 kilometers, and its polar radius is 116 kilometers.

If we plug in the mean radius, we get a surface gravity of 0.048237 meters per second per second; and if we plug in the polar radius, we get a surface gravity of 0.073305 meters per second per second.

If we reverse-figure from 0.0729 meters per second, keeping the stated mass, we get a radius of 116.322 kilometers.

 

I'm guessing the article is using the polar surface gravity; and since the polar surface gravity wouldn't be subject to the same centrifugal concerns, they wouldn't even have to factor that in. I'd still like to know how to calculate Sylvia's equatorial surface gravity though, accounting for centrifugal force.

Thank for the correction. After tinkering with the numbers I came to the same conclusion that you did - Wikipedia is citing the polar surface gravity using the smallest dimension (232 km diameter).

 

Centripetal acceleration would be essentially the same as the "gravity" for an astronaut in a rotating spaceship. To prevent motion sickness in the general population of a space colony (from the coriolis effect) you would want a radius of about 900 m or more with a rotation of about 1 rpm or less if you're shooting for ~1g.

 

Chris

 

Edited to change "diameter" to "radius" and 1 km to 900 m in last paragraph (2*pi*900 m/min*60s/min)=94.248 m/s, and (94.248 m/s)² /900m=~9.87 m/s²

Edited September 3, 2011 by csmyth3025

[Iggy]

Perhaps the figure they give, 0.07 m/s^2, accounts for the potato shape.

 

To calculate equatorial surface gravity with the centrifugal force considered subtract v^2/r from the gravitational force.

[swansont]

If Sylvia's surface rotation speed is greater than her surface orbital velocity, shouldn't Sylvia have a negative surface gravity? If an astronaut wanted to land on her surface, wouldn't he have to use mountain climbing gear just to keep from flying off into space?

 

1. No, never.

2. Yes, but this is not due to 1.

 

Gravity is solely due to the mass of the planet; it is never going to be negative (outward). If you were at the rotation speed given by v^2 = GM/r you would feel weightless, because you would be in orbit at the surface. If v were greater*, you would indeed need to pull down along r to hold on. There is no outward radial force present, if you are solving this in a frame where F=ma holds (i.e. an inertial frame). If you didn't you would be thrown off tangentially — not radially.

 

*the planet would probably not exist under that situation because there would insufficient force holding the surface in place

[Elen Sila]

Perhaps the figure they give, 0.07 m/s^2, accounts for the potato shape.

To calculate equatorial surface gravity with the centrifugal force considered subtract v^2/r from the gravitational force.

 

So, 0.026758–64.7^2/192000=0.004955; at Sylvia's equator, then, the effective gravity would be only five millimeters per second per second?

 

On the earth, (398600.4418*10^9)/6378137^2=9.798285 meters per second per second; but 9.798285–(6378137*pi*2/86164.1)^2/6378137=9.764370 meters per second. The value Wikipedia gives is 9.780327 meters per second per second. The difference is small; but I'm not understanding where it's coming from: 398600.4418 is known to be the standard gravitational parameter of the earth to a very small margin of error; and 6378137 is likewise known to be the equatorial circumference of the earth at sea level in meters to within a very small margin of error; yet I produced a somewhat large margin of error as my result, both with centrifugal force and without it. I don't get it.

[Iggy]

So, 0.026758–64.7^2/192000=0.004955; at Sylvia's equator, then, the effective gravity would be only five millimeters per second per second?

 

On the earth, (398600.4418*10^9)/6378137^2=9.798285 meters per second per second; but 9.798285–(6378137*pi*2/86164.1)^2/6378137=9.764370 meters per second. The value Wikipedia gives is 9.780327 meters per second per second. The difference is small; but I'm not understanding where it's coming from: 398600.4418 is known to be the standard gravitational parameter of the earth to a very small margin of error; and 6378137 is likewise known to be the equatorial circumference of the earth at sea level in meters to within a very small margin of error; yet I produced a somewhat large margin of error as my result, both with centrifugal force and without it. I don't get it.

In both cases the error likely stems from using gm/r^2 for a non-spherically symmetric body. Clairaut's theorem should give you close to exact answers for an oblate spheroid.

[Elen Sila]

Oh, whoa. I just realized. For any given object in keplerian freefall, the semi-major axis of the orbital ellipse is the distance at which centrifugal force exactly equals gravitational force. And, by extension, the gravitational mass of a planet is exactly equal to the sum of the mean orbital distance and the square of the mean orbital velocity of any given satellite of negligible mass.

 

v²/r=gm/r²

gm=rv²

 

As I understand it, astronomers really like it when they find two objects of comparable mass, orbiting each other, because somehow it helps them determine the mass of the system. I would think it would be more helpful to find an object with a negligibly small object orbiting it; why is this not preferrable?

 

[Iggy]

That is probably referring to stars. We can't resolve planets orbiting a single star because the star is so bright, so there is no way to know their mass (other than our sun). With binaries we can observe the velocity of each around the center of mass.

[Elen Sila]

That is probably referring to stars.

 

No, I've totally heard of scientists creaming their pants over binary asteroids or double-dwarf-planet systems like Pluto and Charon.

 

We can't resolve planets orbiting a single star because the star is so bright, so there is no way to know their mass (other than our sun).

 

... Yes we can know the masses of exoplanets. We just measure the back-and-forth motion of the star. How long it takes to move back and forth tells us the orbital period; and how much it moves back and forth tells us the mass.

 

With binaries we can observe the velocity of each around the center of mass.

 

But wouldn't that just tell you the combined mass, as per g(m₁+m₂)/r²? Why would that tell you the ratio between the masses, as well, as I've often heard it's supposed to?

 

OR WAIT. Of course you'd be able to tell the ratio between the masses! You'd be able to see the motion of both of them around the common center of mass; and then you'd just measure the ratio of their distances to the center of mass. Gosh, I'm dumb sometimes. Alright. I see what you're saying. Confusion cleared up.

[Iggy]

Yes we can know the masses of exoplanets.

The mass of the star I mean. I think everything else you asked is answered here: http://outreach.atnf.csiro.au/education/senior/astrophysics/binary_mass.html

[Larry1957]

If the centrifugal force resulting from an asteroid (or planet's) rotatinal speed resulted in a -g situation at it's surface, wouldn't it fly apart. Any loose material on it's surface would certainly dapart. Wouldn't it be something if the asteroid belt was actually the result of a planet who's orbital speed became so great, that it's centrifugal force overcame its gravity? If Sylvia hasn't flown apart, it must be a solid mass. I have a primitave, but unique understanding of centrifugal force. I produced a short video to explain it if you're interested. http://www.scienceforums.net/index.php?app=core&module=global&section=register&do=auto_validate&uid=61243&aid=b33a2dde65fce4d397e5c91cb3f64906

 

Larry

[Airbrush]

Interesting how close your title is to my idea for "enhancing" gravity for humans on the Moon, Mars, and perhaps on the asteroids. I've never heard it mentioned before, except artificial gravity by centrifugal force in zero gravity of outer space. On the Moon or Mars there is some gravity but long-term human habitation would suffer physically from gravity deprivation. It is possible to increase gravity if the crew quarters are located in a slow spinning structure that "adds" gravity by boosting the G force. Where they live would be a giant centrifuge, but the floor is tilted towards center. This is comparable to a Merry-Go-Round. The effect of these living quarters is to add gravity to the Mars or Moon gravity to bring it to a comfortable Earth-like one G.

Edited November 17, 2011 by Airbrush

[Larry1957]

Sorry. Posted the wrong link to my video re: centrifugal force.

http://s1020.photobucket.com/albums/af323/LarryWagner1234/?action=view&current=CentrifugalForceexplaination-1.mp4

 

Larry