Here is the question it came with:  Newton's Law of Gravitation says that the magnitude F of the force exerted by a body of mass m on a body of mass M is:  (Then had the formula above)where G is the gravitational constant and r is the distance between the bodies.
134315
The derivative of the potential function is the gravitational force.\n\nPotential(At distance r) = G*Mass/d\nForce = Derivative(Potential)\nForce(At distance r) = G*Mass/d^2\n\nWhere G is the gravitational constant and r is the distance from the center of the sphere.\n\nSo when you are considering 2 bodies of mass M & m we get the formula (GmM)/r^2.
475187
According to Newtonian gravity, gravitational force is -GMm/r^2.  Any mass experiences a gravitational force toward every other mass in the universe, proportional to the product of their masses, and inversely proportional to the square of their distance.  This means my body experiences a certain gravitational force from the sun, no matter what else is happening in the universe.  That says it would be impossible to shield me from the sun's gravity.\n\nTo find the net gravitational force on me, simply do a vector sum of all the gravitational forces on me from all other masses in the universe.  Simple to say, not so simple to do.\n\nThat sum is zero at the center of mass of the universe, wherever that is.  Unless the expansion of the universe is perfectly uniform, that point will shift with time.  I could find other points where the sum is zero.  I move away from that center of mass but toward a large nearby mass, just close enough that the pull of that nearby mass exactly balances the pull of the rest of the universe.  These will be single points, not regions.  At these points, I still can't say there's no gravitational force acting on me.  But I can say there are countless gravitational forces acting on me, but their sum is zero.\n\nLet's say I tell you I found such a zero-sum spot, about 1/3 of the way between our sun and Alpha Centauri, and I take you there.  How will you know I am telling you the truth?  The principle of equivalence says that there's no test you could make to tell whether the net gravitational force there is zero, or whether you are just accelerating according to the effect of the nonzero net force on your mass.\n\nThe relativistic warping of spacetime in the presence of gravitational mass is more a way of representing or measuring that force, rather than explaining it.  Neither Newton nor Einstein could tell us why gravity exists or how it works.  If spacetime is not warped in your region, it doesn't mean the gravitational field is zero, it means it is uniform.  And again, how would you tell whether spacetime is warped in your region?\n\nI've tried for decades to think of a gravity shield or a gravity insulator.  If I do find one, watch for my Nobel Prize in physics, and for me to quickly surpass Bill Gates on the wealth scale.
43415
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
505024
The gravitation force between two bodies is expressed as\n\nF = G*M1*M2/r^2\n\nwhere G is the universal gravitation constant, M1 is the mass of object 1, M2 is the mass of object 2, and r is the distance between them. The force acts along the direction of r.\n\nFor your case, you need to place the third sphere such that the gravitation force from sphere 1 is balanced by that due to sphere 2. This means\n\nF1 = G*M1*M3/d1^2 = G*M2*M3/d2^2 = F2\n\nwhere d1 is the distance along the x-axis between sphere 1 and sphere 3, and d3 is the distance along the x-axis between sphere 3 and sphere 2. Note that x = d1 + d2, where x = 69.6 m.\n\nSimplifying the above equation yields\n\nM1/d1^2 = M2/d2^2\n\nsubstituting d2 = x - d1 and rearranging gives\n\nM1*(x - d1)^2 = M2*d1^2\n\nM1*(x^2 + d1^2 - 2*x*d1) = M2*d1^2\n\n(M1 - M2)*d1^2 - 2*M1*x*d1 + M1*x^2 = 0\n\nThis quadratic equation is readily solved for the value of d1. Be sure to choose the appropriate root such that it is consistent with the geometry of the problem.
370927
d is the most correct answer, yes.  \n\nHowever keep in mind that despite g being the "gravitational constant", acceleration caused by gravity DOES change depending on the the distance from the earth.  When we're talking about the difference of a table, there is practically no difference, however keep in mind that it does change.
514115
Basically the law you are talking about is the theory that an object in motion will stay in motion until another force acts upon it.\n\nMomentum is calculated by multiplying the mass by the velocity of the object.  There are many applications for these caclulations like if you wanted to find out what would happen in a head on collision between two objects.  You would be able to calculate in which direction the two objects would travel after the collision.\n\nAnyway, here is a website that I found for you that explains it completely.
501881
acceleration due to gravity at a distance 4 times the radius is GM/(4R)^2 Where M is the mass of the earth . that is the answer is g/16.
344185
Plug the numbers into the Newtons Equation :\n\nF =  (GMm) / r^2\n\nF = Gravitational Force / weight  \nG = Universal gravitational constant   6.673*10^-11 (note the - sign on the exponent)\nM = mass of earth (kg)                 5.98*10^24 kg\nm = mass of satellite (kg)             150kg\nR = radius of earth    (m)             6,378,137 m\nh = height above earth (m)             12,800,000m\nr   = h+R\n\nThe answer will come out as the weight being @ 162.67N\n\nNot weightless, although someone standing on the satellite would feel weightless as its in continuous free fall.\n\nEver been in an elevator and felt lighter as it falls or been weighless on the big dipper as it plummets?\n\nThe gravitational force (weight) of 162.67N is what keeps it in orbit and prevents it from flying off at a tangent. Basically it pulls the satellite towards the earth but the satellite is travelling so fast it misses and continues round in its orbit.\n\nAdded:\n\nWeight is measured in Newtons and is the attractive force between the two masses. (M & m)\n\nIn this case the earth and the satellite. A 1 Kg mass on the surface of the earth will be pulled towards the earth's centre with a force of 9.8N, therefore its weight at ground level is 9.8N. If you plug take m as 1kg, h as 0 and plug them into Newton's formula should get the answer 9.8N, which is where we get the constant g from. \n\nAs you get further away from the centre of the mass the gravity field decreases in inverse proportion to the distance squared. Hence the divide by r^2 on Newtons fornula. Therefore the weight of a 1Kg mass at 12,800km is 1.08N.\n\nBecause the satellite is in a stable orbit the opposing forces of the satellite trying to escape orbit and the gravitational pull towards the earth are equal so yes the weight is 162.7N and the centrefugal force is 162.7N. This is the correct answer to the problem. \n\nAs an aside; for simplicicities case in this scenario the mass of the satellite can be regarded as constant due to its relatively low velocity compared to the speed of light. \n\nHowever as you come closer to the speed of light mass increases according to Einstein's theory of relativity, eventually reaching infinity at light speed.
160233
N is Newtons, a unit that measures force. For every 1kg of something, if it accelerates at 1 meter per second per second (m/s/s or meters per seconds squared), than there is a force of 1N acting on it.\n\nForce is calculated by multiplying the mass (not weight) with the acceleration. For example, if you are 75kg, than there is a force of 75kg X 9.8 m/s/s (acceleration due to gravity: roughly constant on Earth) = 735N pulling you down. BUT, there is a reaction force, in this case the NORMAL, pushing you back up at the same time and force. Therefore, the force pulling you down, 735N, take away the NORMAL, also 735N, and you will have a NET FORCE of 0N because you are stationary. \n\nSo in order to resist something pulling at 14N (ignoring air resistance, friction, etc), it needs to pull back 14N.
72828
The measure of force of gravity on an object(6 letters)\n--weight\n\nA unit that measures force(6 letters)\n--Newton\n\nthe rate of changein velocity(12 letters)\n--Acceleration\n\na chnage in position in a certain amount of time \nm_t_ _ n\n--Motion\n\nacceleration is 9.8 meters/sec/sec \n_ _ _ v _t_\n--Gravity\n\na force that is opposite to the direction of motion\n[_]_ _ _ t _ _ _\n--Friction
619208
I assume that the right direction is up as in apposite to the Earth’s gravitational pull.\nWork=Energy = (Force) x (distance)\n\nThe force equals to the weight of the book which I hope still is F=mg=1kg x 9.81m/s^2=9.81 Newtons (N)\n\nThen the distance must be d=Work/Force=1Joul/9.81N= 0.102 meters or 10.2 cm\n\nWas that useful?
377636
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.
92832
This is just an application of Newton's Second Law: F=ma, just remember that acceleration is really the change in velocity divided by time. In the first case, the acceleration is (4.5 m/s)/(0.0024 s) = 1875 m/s^2. Thus the force will be mass times acceleration or 75 * 1875 = 140625 N (1.4*10^5 N). The same logic will get you the 3.4*10^3 N for part B.\n\nFor part C, the net force will be the force of the ground minus the force of gravity (you should draw a free-body diagram to confirm this). So the force of the ground will be equal to the net force (which you already calculated) plus the force of gravity. Gravitational force is going to be mg (m is the mass of the person and g = 9.8 m/s^2). So just add mg to both of your previous answers to get the answers to part C.
412015
we have F= G*M*m/r^2  and F=weight\n\nwhen M=M and r= r/2 then F=G*M*m/ (r/2)^2\n\ntherefore F(new)= 4*F and hence new weight is 4 times.\n\nthis is due to change in the value of g(acceleration due to gravity)\n\nsince F=mg= G*M*m/r^2   and hence only g is effected\n\nand hence weight also changes mass remainig same.
556817
F = -G(m1*m2)/r^2\nF = force\nG = gravitational constant = (6.6742 +/- 0.001)*10^(-11) N * m^2 / kg^2\nm1 = mass of object 1 = 47 kg\nm2 = mass of object 2 = 6.00×10^24 kg\nr = distance between objects\n\nF = -(6.6742 +/- 0.001)*10^(-11) * (47 kg * 6.00*10^24 kg) / r^2\nF = -4.225 * 10^14 N*m^2 / r^2\nSince the distanct between the Earth and the person weren't given this is all the farther the force can be calculated to.  The force will be equal on both the objects (person and Earth) only that the acceleration of the Earth will be almost nothing due to its size and the force will be in the opposite direction.\n\nBut based on the acceleration of the person, you can calculate a force, but it will only be an estimate\nF = m*a = 47 kg * 9.8*10 m / s^2 = 460.6 N\nForce on Earth is 460.6 N\n\nNow the acceleration of the Earth is\n460.6 = 6.00*10^24 * a\na = 7.68 * 10^(-23) m/s^2
368686
The work done is force times length. The force acting in this case is just gravitational force which is also the weight. So you will get \n\nA=M*L\n\nM-weight, L-length, A-work\n\nDon't forget to convert Length to meters.
343167
Very simple, in your case you only need to convert from N(Newton) to Kg(kilogram) so just divide by 9.8 (gravity constant)\n\nN/m2---- /9.8 ---> Kg/m2\nN/m2<-- X9.8 ---- Kg/m2