Gravity keeps us from flying off a spinning Earth. How fast would the Earth have to spin to throw us off?

Question

The Earth spins at about 1000mph at the equator. Since we have a tendancy to follow a straight line, gravity has to counteract to pull us back down. At what rotation rate would we be thrown off the Earth?

Answer 1

for earth,the escape velocity is about 39,600 km/h and the Rotational velocity at equator is 465.11 m/s.and Equatorial surface gravity is 9.780 1 m/s²
(0.997 32 g) .so,.to fling us all and water out of earth,the earth needs to attain these great speeds and even more..but by then..the earth will burn itself dlly destroyed and no life will exist..

hope this helps u..

Answer 2

Because of the size involved, the large-scale behavior of the "solid" matter of the earth is that of a liquid, which takes on a minimum-energy equilibrium shape that results in the total (grav. + centripetal) acceleration vector being perpendicular to the surface at any point. (If it wasn't perpendicular, there would be flow until it became so.) This is the same phenomenon that shapes the paraboloid surface of a liquid in a rotating glass. Currently earth is a slightly vertically-compressed spheroid; i.e. a low-eccentricity ellipsoid. At an initial rotation rate just high enough to cause 0 g at the equator in its current shape, it would assume the shape of a very vertically-compressed spheroid. This would increase its moment of inertia and slow its rotation rate. There would still be poles, and g would be greater there while centripetal acceleration = 0. (Incidentally the initial velocity at the equator is the same as that of a zero-altitude satellite in circular orbit, about 7.9 km/s, not 250 km/s as another answerer thought, and the day is about 84 minutes long.) With a much higher initial rotation rate, mass would be lost but the final shape of earth would be a similar highly compressed spheroid. Mathematicians and physicists have studied the problem. The ref. is a presentation of some of their ideas. I don't know what you mean by a gravitational force potential difference. Yes, grav. force and centripetal force would vary with latitude over a much wider range than they do now, but because of the global perpendicularity of total acceleration at the surface, there would be no large-scale uphill/downhill effect, just as there isn't now.

Answer 3

Near earth satellites travel around the earth in about 90 minutes. If the earth revolved at about that rate then things on the surface near the equator would be thrown off into space. Instead of 1000 miles an hour the equator would be traveling at about 16,000 miles per hour.

Answer 4

The critical condition is NOT to be rotating with the ESCAPE VELOCITY at the equator, but rather with the CIRCULAR ORBITAL VELOCITY.

That velocity's value, which is indeed the velocity at which equatorial break-up would naturally occur, is the escape velocity divided by sqrt(2), or about 17,500 mph in round terms. So the answer is that Earth would have to spin on the order of 17.5 times faster.

To obtain a more accurate figure, simply require that:

v_eq^2 R_e / (G M_e) *** = 1, where R_e = Earth's equatorial radius and M_e = Earth's mass. (I'll let others do this math.)

*** This dimensionless factor is related to a measure of the Earth's non-sphericity, or ellipticity, (R_e - R_p) / R_e where R_p is its polar radius. Very roughly speaking, the numerator here is ~ 13.5 miles, and the denominator ~ 4000 miles, so the ellipticity is ~ 0.00338 or 1/296. Theory says this should scale as the square of the equatorial speed, so gravity could no longer hold objects in at a rotational speed about sqrt (296) greater than it now possesses.

Well, sqrt (296) is ~17.2, confirming yet again the speed-up of rotation we already found above. That means that earth would have to rotate in approximately 24/17.2 hrs or in about 1.4 hours! (Later addition: the "exact" result is now evaluated below. See my POSTSCRIPT.)

(All of this of course ignores such complications as the fact that Earth's own shape, in particular the value of its "equatorial bulge," would change under such rapid rotation, so that the value of R_e to be inserted into the critical parameter would no longer be the current value. The full problem is obviously more complicated and messier that we can deal with here. Nevertheless, the estimates given above are examples of what scientists call a "back-of-the-envelope" calculation of the order of magnitude of a more accurate answer to the question.)

Live long and prosper.

POSTSCRIPT O.K. guys, it's time to get serious:

There are two ways of answering this question "exactly" --- but in the approximation where we stll assume that Earth's equatorial radius will retain its curent value until things fall apart, and the centre cannot hold ... :


1. In terms of the SPIN SPEED OF EARTH'S EQUATORIAL SURFACE, V_e, say.

The critical condition is (V_e)^2 = (r_e) (g_e).

(Check dimensions, using "[...]" for "dimensions of ...": [LHS] = L^2/T^2; [RHS] = L (L/T^2) = L^2/T^2. They CHECK. Note that omegared (below) neglected one "T^(-1)" in his expression for g_e.)

Now evaluate V_e = sqrt {(r_e) (g_e)}.

With (almost!)*** omegared's inputs I obtain V_e = 7894 m/sec or 7.894 km/sec (4.905 miles/sec), very close to the results ~8 km/sec (~ 5 miles/sec) that I always remember for the speed of the lowest possible Earth satellite orbit. Note that 4.905 miles/sec is 17, 660 miles/hour. So this is the "exact" answer to 4 sig. figs., in this linear "speed at the equator" way of expressing it.

*** Why was omegared's result so different? Because Earth's radius was entered as 6372 METRES instead of 6372 KILOMETRES, an input error of 10^(-3).


2. A second, entirely equivalent way of expressing the result would be to evaluate the critical value for the Period of Rotation,
which is in fact the same as that of the lowest conceivable satellite orbit, assuming a negligibly thin atmosphere.

While one could use the value for V_e already determined, along with r_e, a more self-contained and aesthetically satisfying way to do it is to use another equally valid formula:

P_crit = 2 pi {r_e / g_e)}^(1/2)

[Note: It is NO coincidence that THIS is "the period of a pendulum with the extremely long length r_e in Earth's surface gravity field, g_e." The proof is left as an exercise to the student!]

O.K., let's evaluate P_crit. It's:

5072 secs = 84.53 mins.

Because P is proportional to r^(3/2) (remember Kepler's 3rd law), this is the SHORTEST POSSIBLE, "EARTH-GRAZING" period that an Earth-satellite can have. Put the satellite up at an elevation of 200 kms, and the period would be ~ 88.5 mins. That's why the "90 mins" approximation is a fairly good one, in practice.

To get back to the question at hand:

Rotate the Earth itself with a period of 84.53 minutes, and we wouldn't be thrown off the Earth, but would rather simply feel we were "floating," apparently "weightless" at the equator. If we could be impulsively accelerated to a rotation rate faster than this, then we would indeed be "thrown off the Earth," into an elliptically grazing orbit having perigee at potential "touch-down" points on the equator successively further and further west with the completion of each "orbit." (In other words, we would see the land/sea immediately below us drift slowly to the east.) This is NOT necessarily a recommended route to adopt in order to obey the old dictum of "Go West, young man!"

Answer 5

Funny, I had that question on my Physics final test. I remember that if people want to stay afloat on the equator(counteract their weight), the Earth has to be spinning fast enough so that one day will be a little two hours (like one hour and 56 minutes or something).

If you want people flying off, then anything faster than that would work. The faster it spins, the faster people fly off. The Earth will bulge quite a bit on its equator anyway and countless other effects which would kill us all off before we fly off.

We (espcially humans) are very fragile and the environment we need is VERY specific and narrow ranged (such as temperature, gravitational force, air pressure, etc...) Any little deviation can b devastating.

Answer 6

The escape velocity from earth, as I recall, is 25,000 mph. The earth is now spinning at the equator at about 1,000 mph (the earth is about 24,000 miles around and goes round in 24 hours.) So to fling us and water, etc., off it would have to spin about 25 times as fast or roughly one revolution per hour.

Answer 7

As long as the gravitational acceleration is greater than Centripetal acceleration we'r safe

gravity is 9.780 m/s
so for :Centripetal acceleration>gravitational acceleration
V^2/r>9.78
V^2>9.78*6372 (r=6372)
v>250m/s

currently earths spin is abt .5m/s

Answer 8

What Mike didn't mention is that the Earth would break up long before it reached that speed. . .