Universal Gravitation physics problem.?
969045
Are you taking into account the radius of the moon? The equations you've provided treat "r" as the distance between the centers of mass of each object. Therefore:\n\nr = radius of the moon + distance from it's surface\n\nr = 1785 km + 111 km\n\nr = 1,896,000 m\n\nSubstituting this into your equations provides:\n\nT = 2*pi*sqrt((1896000)^3/(6.674x10^-11 * 7.3x10^22))\n\nT = 7431.6 s = 123.9 min\n\nAdditionally:\n\nV = sqrt((6.674x10^-11 * 7.3x10^22) / 1896000)\n\nV = 1603 m/s\n\nHope that clears things up a bit! :)
904747
Its origins and nature are still open to postulation. But it is an attractive force only between any two masses and Newton concluded in his uneiversal laws of gravitation that the attractive force (F) was proportional to the masses interacting(M1 and M2) and inversely proportional to the square of their distance apart (r)\n\nie F α M1*M2/d²\nwhence \n\nF = G*M1*M2/r² \nwhere G is the universal gravitation constant\nG ≈ 6.6742 x 10^(-11) Nm²/kg²
771534
The problem is about conservation of angular momentum.  The tricky part is (b). To solve this use this definition of angular momentum and draw a picture to find out the value of r sin θ\n\nThe angular momentum of a particle of mass m with respect to a chosen origin is given by vector product L = r x p= mvr sin θ
447698
Surface gravity 273.95 m s-2\n                 (27.9 g)
207775
Mass is only one variable in an equation... what are you trying to relate it to?\n\nweight = mass x gravity would be one equation for mass...
384756
Yup, relativity and qunatum mechanics work together nicely.  You'll need to account for the switch to using spacetime quantities instead of space and time quantities, and the fundemental equations get kind of hairy, but they still work out fine.
757859
haha, k  Draw a picture and draw a free-body diagram.\n\nI don't feel like busting out a paper/pencil  so I'll try it in my mind\n\nF(sub g) cos Theta  =  Mg cos(20 degrees)\n\nbecause the Normal force is equal to the y-component of gravity.\n\nTwo similar triangles, the incline and one which has the Force of gravity straight down, and the y compentn equal opposite of the normal force, and X component equal opposite to F\n\n\nlots of physics i'm skipping in my explanation tho
939750
W=mg ... where g is the gravitational acceleration
950533
You must be talking about the planet called Venus. \n\nto calculate the acceleration due to gravity, you have to use the following formula:\n\ng = GM/(r^2)\n\nwhere G is the universal constrant of gravitation = 6.673x10^-11\n\nyou know that:\n the mass (M) is 4.88 x 10^24 kg\n and the radius (r) is 6.06 x 10^6 m\n\nnow substitute into the equation. \n\ng = [(6.673x10^-11)(4.88 x 10^24)] / [(6.06 x 10^6)^2]\ng = (3.256x10^14) / (3.672x10^13)\ng = 8.867m/s/s\ng = 8.87m/s/s (thats metres per second squared)
974496
Equatorial surface gravity \n1.35 m/s²\n0.14g
61784
That's an interesting qts. I assume that you wish to invlove pure physics in this qts. Forget all the heat at the core of the earth and also air resistance.\nConsider pure gravitation theory.\n\nIf say the fellow who fell in the hole has mass m\n\nThen force of gravity exerted by the earth is :\n\n  F = GMm/(r*r) ..... Where G - universal gravitation constant\n\n                            M - Mass of the earth\n                            r - distance between earth and body of mass m\n\n          acceleration due to gravity\n            = (1-h/R)g   .... where h is distance of body beneath the earth's surface and g is acceleration due to gravity at the surface of the earth.\n\nSo as the body comes closer to the core of the earth, h tends to R.\nSo the acceleration of the body goes on decreasing very much. So, body may not actually hover, since his velocity will be constant the instant it is at the core.
603793
Anew theory of gravity as well as a unified field theory (UFT) are being proposed, showing how the four force fields work together, implicating also a fifth field. A main feature of this new model is the proposal of the existence of diallel, gravitational-field lines that emanate from every entity. For the earth, for example, they emanate radially from its center. These diallel lines provide a conduit for combined particle and wave flow and exchange beyond the speed of light, and they have quantum states similar to the atom. On the web site, we present a general field equation showing the importance of energy density in this new UFT. We also present a new gravitational force equation, that has as a special case the commonly accepted gravitational equation.
967250
The gravitotional force F of two masses m1 and m2 separated by a distance r is given by the equation\n\nF = G * m1 * m2 / r^2\n\nwhere G is a physical constant with value 6.67*10^ -11 N*m^2/kg^2\n\nSInce the third object is placed midway then its distance from either m1 or m2 is 0.15 m.\n\nSubstituting we get\nF1=3.9*10^-5 N\nand\nF2=9,5*10^-5 N\n\nSince the two forces have opposite directions then \nFtotal = F1 - F2 = -5.6*10^-5 N having the direction of F2.\n\nIn order for total to be zero we have\nF1=F2 ->\nG*m1*m/x^2=G*m2*m/y^2 ->\nm1/x^2 = m2/y^2 ->\nm1*y^2=m2*x^2 (1)\n\nwhere x and y the diastances of m from m1 and m2 respectively.\n\nsince\n\nx+y=r=0.300 m\n\nit follows that y = 0.3-x (2)\nCombining (1) and (2) we have\n\nm1*(0.3-x)^2=m2*x^2 ->\n215*(0.09-0.6*x+x^2)=515x^2 ->\n300x^2+129x-19,35=0\n\nsolving this we get\n\nx1=0.12m\nand\nx2=-0.55 m which is of course rejected since it represents a point where there can be no equilibrium.
987930
No energy source? What source of energy is needed? And the simple presence of mass causes gravitons. And these gravitons interact with one another. And proof, you need proof that gravity works? Dude go jump off of a building and see what happens!
1053598
http://en.wikipedia.org/wiki/Gravitation#Newton.27s_law_of_universal_gravitation
205433
At this point the big thing we're looking for is the higgs boson.  until we find that, all are ideas of what gravity is are unconfirmed.  All we can do is describe how it behaves.
125809
The struggle here is that you provide too little information to recognize this problem as coherent.\n\nWe may assume that the acceleration of gravity is 9.8 m/s/s or 32 ft/sec/sec.\n\nThe velocity(V) of an object is equal to its initial velocity(v) plus its acceleration(a) times time(t). \nV = v + at or V = v - 32t (in feet per second)\n\nYour object will reach maximum height when the velocity is zero or when the initial velocity is equal to 32 ft/sec/sec * time. With your list of times, you'd just multiply by 32 to solve for the initial velocity, if that is what the problem is actually looking for. There are many other permutations of this data, but hopefully that is a start for you.\n\nGood luck.