How do you solve for gravity?
100886
The equations weren't visible in Ginabean's answer (at least not on my display) and the other answer dealt with Earth, not the fictional planet postulated.  So I'll add some things.\n\nWe worked out the mass of this planet using the density and the radius in your other question.  M = 8.92*10^22 kg.\n\nNow, the force of gravity is given by\n\nF = G M m /r^2  , where M is the mass of the first object, in this case the planet, m is the mass of the other object, in this case something on the surface of the planet, and r is the distance between their centers, in this case the radius of the planet.  (Because the planet is spherical, we can treat the mass of the planet as if it were all located at its center.)\n\nSo, the acceleration caused by gravity on this planet, g, is F/m where F is the force of gravity and m is the mass of the object, \n\ng =F/m = G M /r^2\n\ng = (6.67 * 10^-11 m^3/s^2kg) * 8.92*10^22 kg / (1600000 m)^2\n\ng = 2.32 m/s^2\n\nThis is about a fourth of the values on earth.  That makes sense.  The mass and radius of this imaginary planet are comparable to the Moon (M = 7.36 x 10^22 kg, r = 1738 km) where the acceleration of gravity is about 1/6 of the Earth.
630060
Speed of.. gravity? :S
1045412
gravity is an inverse square law type graph , you can plot it at y=1/x^2\ntry this site http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html\n\nanother graph is available here\nhttp://csep10.phys.utk.edu/astr162/lect/light/intensity.html
869887
If you have 2 equations, just solve for one variable for one of the equations. Take that equation and plug it into the same variable for the other equation. You will solve for that variable, and plug in that number to any of the 2 equations to solve for the other variable.
899235
A) I think I got it...Use energy, except for g, use   Fg cos(19 degrees), found from a free-body diagram.\n\nso 1/2 mv^2 = mgh\n\n1/2 (4.7)^2 =  9.8(cos (19))  h    solve for h\n\nB)  now that you have h...maybe you can use   \n\nd y = V(sub 0 sub y) t + 1/2 (a x cos (19)) t^2\n\nyea, tell me how it comes out.
823207
gravity is a special case  of gravitation.It is the force by which earth pulls a one kg object towards it.\n\nie g=GM/R^2\n M=mass of earth\nR=Radius of earth\n\nnow force acting on a object  of mass 1 kg at a height h(from earths surface) will be \nf=GM/(R+h)^2\nwe can take out R out of bracket so that (R+h)^2 can be R^2{1+h/R}^2\nNow\nGM/R^2=g\nSo we can easily WRITE it as \nf= g/{1+h/R}^2\n\nThis f is that something we usually write as g'\n\nNow g' can be written as g{1+h/R}^(-2)\n\nFOR NOT TOO SOPHISTICATED CALCULATION\nIT CAN BE REDUCED USING BINOMIAL THEORM TO \ng'=g{1-2h/R}
997326
Use Newton's universal gravitation equation, then sum the forces.\n\nF = G*m1*m2/r^2\n\nF = force (newtons)\nG = gravitational constant (6.6742 x 10^-11)\nm1 = mass of object 1 (kg)\nm2 = mass of object 2 (kg)\nr = distance between objects (meters) ==> don't forget to convert the units\n\nIn this case, you do it twice:\nF1 = G * 390 * 390 / 0.11^2 = 0.84mN to the left\nF2 = G * 390 * 390 / 0.39^2 = 0.067mN to the right \n\nNet Force = 0.84 - 0.067 = 0.772mN to the left
945010
Equatorial gravity 3.69 m/s2\n(0.376g)
374428
The most strait forward way is to use the quadratic formula. You just do simple adding, powering, rooting, and you get the answer. It works the 100% of the time. There is also the filling-in-the-square, but that's too much work compaired to the quadratic formula.
109780
As you move "downward" towards the center of the Earth, gravity will decrease. As you pass through the exact, physical center of Earth, gravity will be zero. As you move away from the center, gravity will begin to increase.\n\nYou can roughly compute the force of gravity with this formula --\n\ng = G*m/r^2\n\ng - gravitational force in Newtons\nG - gravitational constant (6.673^minus 11)\nm - mass of Earth\nr - distance from center of mass
161857
Solve for the mass of the planet.  You are given a diameter and are told that it is sphereical, so solve for the solume.\nV = 4/3 pi*r^3\n\nOe you know the volume, using its density, solve for the mass,\nD = m/ v\n\nFind the orbital distance of the moon from the planet's center of mass.\nThis distance, r, would be the moons distance above the planet's surface, plus the planets radius.\n\nFind the grvitational acceleration caused by the planets mass on the moon using Newtons law of gravitation,\ng = G*m/r^2,\nwhere m is the mass of the planet, r is the distance from the center of mass, and G is the Universal gravitational constant (6.67 E-11).  Solving this will give you g, the gravitational accelerating the moon experiences towards the planet.\n\nWe are told the orbit is perfectly circular, so we can solve for the orbital speed of the moon by setting the centripetal acceleration eqaul to the gravitational acceleration.\ng = v^2/r\nwhere v is the orbital speed and r is thr radius.
559516
Force of gravitaional attraction is given by:\n\nF=G*m1*m2/(r^2)\n\nwhere:\nF=force\nG=universal gravitational constant\nm1=mass of first object\nm2=mass of second object\nr=distance between objects\n\nr must be infinitely large before force of attraction becomes zero.
747846
An inverse relationship is given by the formula\n\nxy=k\n\nwhere x and y are variables, and k is a constant. For our data, let i=intensity of gravitational field and d=distance. Then a formula for our relationship is\n\nid^2=k\n\nWe know that when d=2, i=6, and use this information to solve for k:\n\n6*2^2=6*4=24\n\nSo k=24. Now we want to know what another i is when d=3. Plug in the information we know:\n\ni*3^2 = 24 -> 9i = 24 -> i = 24/9 -> i = 8/3\n\nThus, the intensity is 8/3. This makes sense since the greater the distance, the less the intensity.
797830
Keep in mind these equations\n\nx(t) = x-initial + (v-initial)t + (1/2)(a-gravity)t^2\nv(t) = v-initial + (a-gravity)t\n\n(v-final)^2 = (v-initial)^2 + 2g(x-final - x-initial)\n\na-gravity = -9.8 m/s = -32 ft/s (i think..)\n\nTherefore\n\nx(t) = 0 + (64)t - (16)t^2\nx(t) = -16t^2 + 64t\n\n(v-final)^2 = (v-initial)^2 + 2g(0)\n(v-final)^2 = (v-initial)^2\n\nsince it is going in the opposite direction,\nv-final = -(v-initial)
975783
The force balance says that\n\nSFx = 0 = -2*g*sin(q) + m*v^2/r\nSFy = 0 = 2*m*g*cos(q) - m*g\n\nwhere m is the mass, v is the speed, g is the gravitational acceleration, q is the angle that the ropes makes with vertical, and r = L*sin(q) is the radius of the orbit. SFx and SFy represent the sum of the forces in the x- and y-directions, respectively.\n\nSFy indicates that\n\n2*m*g*cos(q) = m*g\n\nor\n\ncos(q) = 0.5\n\nHaving the angle, you can solve SFx for the velocity\n\n2*m*g*sin(q) = m*v^2/r\n\nor\n\nv^2 = 2*g*L*sin(q)^2\n\nGiven the velocity, the angular velocity is\n\nw = v/r.
787952
The force of attraction is given by Newton's law, F = G* m1*m2/r^2, where G = universal gravitational constant, 6.67 * 10^-11 N*m^2/kg^2.  You have the two masses, and the radius of earth is the distance r.  Your weight is the same as the gravitational force.
9301
ok, well the formula you need is:\n\nFg=Gm1m2/r^2\n\nwhere Fg is the force of gravitation, G is the universal gravitational constant(just a constant, you dont have to worry about it) m1 and m2 are the masses of the 2 objects, and r^2 is the distance between the two objects squared.\n(a) ok, so if r increases by 10 times, since r is squared, the denominator would actually increase by 100 times.  Since the denominator increases by 100 times, and the numerator stays the same, you're basically multiplying it by (1/100) so that means that you have to do the same thing to both sides, so you just multiply Fg by 1/100 so thats 500/100 so the answer is 5 newtons.  This makes sense because the farther objects get from each other, the less each other's gravity will affect them.\n\n(b) ok, now one of the m's, lets say m1 doubles.  The rest stays the same again, and this time its in the numerator, so you're just multipying the whole thing by 2, so the new force would be 1000 newtons.  This makes sense, because something with more mass would experience more gravitational force, for example, a person with more mass experiences a greater gravitational force with the earth, and therefore weighs more.