Gravity Question. (more)?
955246
If you were right at the center then the symmetry of the situation would mean that you'd be weightless since the gravitational force would be balanced all around.  \n\nSimilar situation if you were just a little off the center.  Most of it would balance except for a little bit.  Think about starting at the earth's surface and moving to the center.  You'd start at your weight and become zero at the center.  Assuming continuity and noting the nice behaviour of gravity then it must decrease in a smooth fashion.  If you continued through your hole to the opposite side you'd weigh the same but in the opposite direction, always pointing towards the center.
546779
I think it is more accurate to say we can easily replicate the EFFECTS of gravity, rather than gravity itself.  Gravity relies on mass.  More mass, more gravity.  Any "replication" of real gravity is not replication it is the real thing.
1119730
ahh, good theory problem...\nlet's see, there's energy, friction, force (normal, tension, as well as the other ones mentioned), gravitation, mass, pendulum motion, gotta be more that I've missed, but no doubt someone'll be here to correct me.
447698
Surface gravity 273.95 m s-2\n                 (27.9 g)
901635
1\na.  distance, slower\nb. weaker gravitational pull farther away\n2\na. 750.353 km\nb. 4.67 cm\nc. Not enough info\n\nPS please award best answer
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
596583
Mass. I think.Depending on how strong the gravity is your mass changes.
812295
Er, gravity really has nothing to do with this question. Fk = coefficient*Fn.  Fn = mg.  -19.6 is your Normal Force.  But yea, the answer is correct.
450131
As far as I know, they are still not sure as to what creates gravitational force. If they determine the source, one might negate it to experience zero gravity anywhere.
474542
Gravity does not change the mass of an object, only its weight.\n\nIf you travel away from the Earth, or any concentration of gravity, you will weigh less, but your mass is unchanged.
324067
No. Gravity has to do with the mass of objects--you have some too.
912569
I think F is the answer. Taking acceleration due to gravity is constant at 9.81ms^-2, but this assumes that air resistance is neglected, and g is a constant.\n\nIf air resistance is NOT neglected, and acceleration due to gravity is not taken to be constant, H is the correct answer. The rest of the options are wrong.
247359
Mass interacting with gravity
365770
i dont' know what causes the rotation of an orbiting planet, but the moon would definitely have gravity even if it didn't spin.  that dude was full of crap.  if anything, the spin would cause things to fly off.
840601
(a) Gravity exerts a downward force on the block of 9.8 Newtons/kg * (0.25 kg) = 2.45 Newtons. Work done by gravity = force * distance = 2.45 Newtons * 0.12 m = 0.294 Joules\n\n(b) The spring is a bit tougher since the upward force increases as the spring is compressed. Force = 250 N * (C) where C is the distance the spring has been compressed (in meters). Work is then the integral of Force over distance, so Int(250 * C (dC)) from C = 0 to 0.12 M . The integral is 125 * C^2. Evaluation at 0 gives 0. Evaluation at 0.12 gives 125 (0.12)^2 =  1.8 Joules. So the work done by the spring is -1.8 Joules . Note that the work is negative, because the object is moving against the applied force.\n\n(c) When the block strikes the spring, it will have a kinetic energy of (1/2)mv^2 = (1/2)(0.25kg)v^2 = 0.125v^2.\nThis kinetic energy is absorbed by the spring as it compresses. Note that the work done by gravity is also absorbed by the spring.\nTherefore, we have the following equation:\n0.125v^2 + 0.294 = 1.8\nv^2 = 8(1.8-0.294) = 8(1.506) = 12.048\nSpeed of block just before it hits = SQRT(12.048) = 3.47 meters/sec downward.\n\n(d) I'll leave this as an exercise for you. Just apply the same techniques as above. Note that the block hits the spring with a velocity of 2(3.47) = 6.94 meters per second...
110744
Short answer: usam da inverse square law Jar Jar!\n\nLong answer:\n\nNewton simplified things by picturing all of a planets mass gathered in a single point in the center of the planet and the surface as a point on a certain distance from that center of mass. So the gravitationall pull (F) on a body (m) at a distance (r) from the mass (M) can be calculated using this equation:\n\nF=G*((M*m)/r^2)\n\nwhere G is the gravitational constant (6.67 * 10^-11 Nm^2kg^-2)\nEarths (equatorial) radius is 6.378 * 10^6 meters and its mass is\n5.974 * 10^24 kilogrammes\nSo lets test this on a dude weighing 100 kg:\n\n6.668 * 10^-11*((5.974*10^24 * 100)/4.068*10^13)\n==>\n6.668 * 10^-11*(5.974*10^26 /4.068*10^13)\n==>\n6.67 * 10^-11* 1.424*10^13 \n=\n979 N (what I got in my calculator)\n\nSo earth pulls on a guy, with the mass 100 kg, with the force of 979 Newtons. That means 9.79 N/kg, earths gravitational force (g). This should also mean that a mass dropped from a small height should accelerate at the rate of 9.79 m/s^2. It´s a simple experiment to do but where I am the value of g is actually 9.82. This is because I am closer to the earths center. Earth bulges out at the equator because of its rotation. I am farther north. But still a simple experiment can be done to validate this calculation. \nSo if the earths mass was 3 times greater than g shoud be 3 times greater. Right? You betcha. And I have already shown that moving closer to earths center increases g. Moving twice as far from the center should decrease g by half. Right? Wrong! Now what? We where so close. Oh man...\ng is reduced to a quarter if we increase the radius by 2. This is because of the inverse square law, a very important law in astronomy. It means that the apparent brightness of a light source is inversely proportional to the square of its distance. This also aplies to gravitational fields. Double the distance, divide by 4.\nSo lets put this all together. You should be able to do it yourself by now. I got 734 N or 75 % of the dudes weight on earth. And the reason he doesn´t weigh more than on earth is because of the inverse square law. Eventhough the new world is three times as massive we still moved twice as far away from the center. So take the weight of your astronaut, multiply by 3 and divide by 4.
30290
draw the free body diagram. ok, there's a force of 21N going down and an unknown force going up. you know that the net force is 12N. so that means how much of the 21N force is being cancelled out by the buoyant force to produce the new weight? so it's just 21N-12N which is 9N