0
multiple family residential
for triplex to 8-plex
Wiki User
∙ 6y ago
Add your answer:
Earn +20 pts
A 75 kg person is running at 2.5 ms along the outside of a stationary merry-go-round when they jump on the outside rim The merry-go-round is a 75 kg disk of radius R2m What is the angular speed of the?
It appears that the end of this question has been cut off. One cannot calculate the angular speed if the object for which it is to be calculated is not specified in the question.
At what height from the earth 'g' value becomes 0?
First we have to be careful because 'g' is defined as the nominal acceleration due to gravity at sea-level on earth. So g, by definition, is always 9.80665 m/s2.Now if we ask "At what height from the earth does the acceleration due to the gravity of the earthbecome 0?" then the "classical" answer, from Newton's Universal Law of Gravity, is infinity (but read the last note at the end of this answer). The classical equation for this acceleration is:GMe/r2where G is the universal gravitational constant (~ 6.67428x10-11 m3kg-1s-2)and Me is the mass of the earth (~ 5.9736x1024 kg)and r is the distance from the earthSo as r increases the acceleration decreases rapidly. This is a prime example of an "inverse square law". At very large distances the acceleration still has a very small but non-zero value, tapering to zero at infinity.Another question would be "At what height from the earth does the total acceleration due to the gravity of all other objects become 0?". The answer is that it depends on what direction you are considering. Directly between the earth and the moon for instance, then if we ignore the effects of other objects (sun, planets, stars, ...) we can calculate the point where the effect of gravity from the earth and the moon "cancel each other out", giving a net acceleration of zero:GMe/r2 = GMm/(D-r)2 Where r < Dwhere Mm is the mass of the moon (~ 0.07349x1024 kg)and D is the distance between the earth and the moon (~ 3.78x108 m)First, simplifying the equation (by dividing by G and multiplying r2 and (D-r)2) :(D-r)2 = r2M (where we'll simply define the mass ratio M = Mm/Me ~ 0.0123)Expanding:D2-2Dr+r2 = r2MTurning into the quadratic equation:(1-M)r2 -2Dr + D2 = 0Solving the quadratic equation:r = 2D +/- (4D2 - 4x(1-M)xD2)-2 / 2(1-M)Simplifying:r = D +/- (D2 - (1-M)xD2)-2 / (1-M)r = D +/- (D2 - D2 + MD2)-2 / (1-M)r = D x (1 +/- M-2) / (1-M)Now filling in the values:r = 3.78x108 x (1 +/- (0.0123)-2) / (1 - 0.0123)r = 3.78x108 x (1 +/- 0.1109) / 0.9877r = 3.83x108 x 1.1109 or 3.83x108 x 0.8891r = 4.25x108 m or 3.41x108 mNow we can throw away the 1st value because our condition at the beginning was r < D (the magnitudes of the two forces are equal at this distance too, but they are both now in the same direction and so add together rather than subtract to zero.)So r = 3.41x108 m = 341,000 km ~ 211,889 miles (or about 90% of the distance to the moon)To put it in perspective that's roughly the total distance you drive in a car over 10 years.OK so what about directly between the earth and the sun (ignoring moon etc.)?A quick flip of the values:D = distance between earth and the sun = 1.496x1011 mM = ratio of sun's mass to earth's mass = 333,000r = 1.496x1011 x (1 +/- (333,000)-2) / (1 - 333,000)r = 1.496x1011 x (1 +/- 577) / -332,999r = -4.49x105 x 578 or 4.49x105 x 576r = -2.60x108 m or 2.59x108 mHere then r = 2.59x108 m = 259,000 km ~ 160,935 miles (or only about 0.17% of the distance to the sun!)Note that this height result for the sun is less than the height for the moon. So how about when the sun and the moon "gang up", combining their gravitational forces against that of the earth. This would happen at the time of a solar eclipse, with the moon directly between the earth and the sun.For this case:GMe/r2 = GMm/(Dm-r)2 + GMs/(Ds-r)2 Where r < Dmor 1/r2 = Mme/(Dm-r)2 + Mse/(Ds-r)2This will produce a 4th order polynomial to solve for those interested to try!Another interesting thing to note is that we should be careful when using this application of the classical Newton's Universal Law of Gravitation further out in the universe. From the discoveries of both the faster than expected speed of stars around galaxies and the unexpected acceleration of the universe itself, it seems there are other factors at work. They may be the entities tentatively called "Dark Matter" and "Dark Energy" respectively or they may be an indication that the Law of Gravitation is only an approximation that works well within the scale of our solar system but requires adjustment at the much larger scales of galaxies and the entire universe.
