Does the acceleration due to gravity on a particle near the earth also depends on the particle's mass?
451065
Yes. The acceleration due to gravity on a particle depends on the mass of the particle.\n\nBecause F=(G.M1.M2)/r^2.\n\nwhere M1 can be considered as mass of Earth.\nAnd   M2 is mass of the particle.\n\nAs the mass of the particle is very small compared to that of Earth, the equation becomes\n\n   F= G.M1/r^2\n\n   Where mass of the particle is neglected.\n\nBut when the particle is too big, like asteroid or other planet, its mass cannot be neglected.\n\nAnd its acceleration towards Earth becomes more than 'g' due to asteroid's own gravitation force towards Earth. Which further accelerates it.
913200
Both their accelerations will be the same and equal to the acceleration due to gravity, g=9.81 m/s/s...This is coz the only force acting on them is the force due to gravity
208514
Yes, anything with mass generates gravity.  Gravity will only effect things which have mass, so the fact that you are stuck down means you have mass, and thus gravity.
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.
324067
No. Gravity has to do with the mass of objects--you have some too.
127639
the acceleration of an object is dependent upon both force and mass...Forces cause acceleration\n\nit takes an action force to cause acceleration for the object\n\naccelerate if there is an unbalanced force acting upon it.
131522
it depends on how you orient your coordinates, but the gravitational acceleration opposes the velocity of the upward traveling ball, so if up is positive then yes the acceleration due to gravity is negative
260158
Suppose you are accelerating due to the presence of some mass M and you are some distance r away from the mass. You yourself have mass m. Then we know that your gravitational potential energy is equivalent to your maximum kinetic energy. Thus GMm/r = (1/2)*mv^2, or v = Sqrt[2GM/r]. If you start far enough away, you can take r to be the radius of the mass you are accelerating towards and then v will be your velocity as you hit the ground, not accounting for relativistic effects. \nIf you are in a starship and using some sort of propulsion relative to some inertial space-time frame, you have an asymptotic upper limit of c as measured from any such  frame, from basic relativity theory.\nIf you are not moving relative to an inertial frame (ie., you are using a gravity drive as described in the source link), then your "velocity" is not limited by any theory in contemporary physics.
425701
No. You can understand the difference by going back to its source: the Newton's Second Law.\n\nF = ma\n\nThe weight of an object is the attractive force (F) exercised by the Earth on that object.\n\nAs the above equation shows, the mass (m) of an object is the constant of proportionality that links any force exercised on the object with the consequent acceleration (a).\n\nSo you see that the weight is a special case of force, that you have when a is the Earth's gravity acceleration. \n\nNow suppose you bring the same object on another planet, say Mars, where the gravity acceleration is different. Its mass doesn't change, but the weight does!\n\nThe weight of an object is an Earth accident, while its mass is a more fundamental concept because its value remains constant (at least within the non-relativistic limit, but this is another story).
804445
First you want to know the acceleration due to gravity of the planet Glob. At the surface the gravitational acceleration (g) can be determined by equating the weight of an object with the gravitational force law\n\nm*g = G*M*m/R^2\n\nwhere R is the radius of Glob, M is Glob's mass, G is the universal gravitational constant, and m is the mass of the object. Solving for g yields\n\ng = G*M/R^2\n\nNow, throwing an object into the air, and neglecting air resistance, we can consider conservation of energy. The kinetic energy of the object when it is released by the astronaut is completely converted into potential energy when the object reaches the top of its travel.\n\n0.5*m*V^2 = m*g*H\n\nwhere V is the initial speed of the object and H is the height to which it traveled, and g is the gravitational acceleration solved for above. Solving for V we find\n\nV = sqrt(2*g*H).\n\nThe satellite traveling in a circular orbit of radius D above the planet has its centrifugal force balanced by the gravitational force of the planet.\n\nFc = m*v^2/D = G*M*m/D^2 = Fg\n\nSolving for the speed of the satellite (v) needed in order to maintain orbit at radius D yields\n\nv = sqrt(G*M/D).
168727
Sort of.\n\nIf you drop an apple then the kinetic energy of the apple has when it hits the ground is proportional the Earth's gravitational acceleration (9.8 m/s^2). \n\nSo if you're on a different planet the "g" for that planet will be different and the kinetic energy of the apple is different, by a proportional amount.\n\nThe same is true of gravitational potential energy.
364601
If that direction were opposite the direction of the \nearth's rotation, I think the answer is theoretically yes.\nIn reality there's not enough mass of people relative to the mass of the earth to have any effect.
297729
You are correct, but the effects are so insignificant, it doesnt nearly matter.\n\nAll object in the universe have a gravitational attraction to one another.\nTake the Earth and bowling ball example,\nThe Earth exerts a force on the bowling ball, and the bowling ball exerts a force on the Earth.  Both on these forces cause the other object to accelerate towards a common center of mass.  In the case of the Earth's gravitational force on the bowling ball, the force is huge in comparison to the bowling ball's force exerted on the Earth, so much so that we can easily neglect the 2nd force.\n\nLets look at some numbers.....\nIf we say the Earth has a mass of 5.9742 E24 kg and say we will say we have a very heavy bowling ball with a mass of 1000 kg, then we can calculate the gravitaional force between them if we know the distance.  Lets estimate their centers of mass are 6380 (~larger than radius of Earth), we can calculate the gravitational force to be,\nF_g = (G*M*m)/r^2\nF_g = 9789.58 Newtons, which would cause an acceleration of 9.7895 mm/s^2 for the bowling ball and a miniscule 1.63 E-21 m/s^2 for Earth.  To put that in perspective of just how slow acceleration is, if you were to accelerated you car from at a constant acceleration of 1.63 E-21 m/s^2, it would take you about 86500000000000 YEARS to go from 0 to 10 miles per hour!  I think for simplicity sake, Newton did all right.\n\nEDIT:\nYes, the bowling ball very much so does pull on the Earth.  How is the "universe" to know which object is suppose to be the planet and which is the bowling ball?  ALL objects in the universe have a gravititational attraction to ALL other objects.  The Earth pulls in the bowling ball, and the bowling ball pulls on the Earth.  There are 2 equal forces pointed in opposite directions, therefroe momentum is conserved (which is a physical LAW).
345655
Gravity is also like another force. it follows Newton's 3rd law. Earth attracts us and we attract Earth.\n\nBut a = f/m. Since our mass is less, we move with greater acceleration. For earth, mass is huge and the effect is negligible.
364646
Acceleration due to gravity on the Earth is 9.81 m/s^2 or 32.2 ft/s^2.\n\nUnless you mean the acceleration of the Earth?  If you assume a circular path (not true, but I'm guessing you're not doing rocket science, so I won't worry too much), acceleration on the earth by the sun would be the square of the velocity at which the eart travels divided by the distance from the earth to the sun. a = V^2/R.
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.
841193
It's the fall velocity at which air resistance exactly balances gravitational force, so the falling object no longer accelerates. However, it varies with air density, which changes as the falling object's altitude declines. Gravitational force also changes with altitude because it depends on the distance to the earth's center. As a result, a falling object is hardly ever going to fall at exactly its terminal velocity, since it has to "chase" the changing value of that velocity.