If the atractive acceleration comes from center of the earth(a point) then at surface ,it s per meter^2?
458068
Pressure is force over area and would be per m^2.  They use the center of the earth because the math works out such that all the mass of the earth could be treated as being at that point.  It doesn't have as much physical meaning as you might think.
150657
gravity is highest at the centre of the earth. so you will be accelerating towards the centre.since there is no resistence (assume so that there is no air), you would not stop at the centre due to inertia of motion and continue to go beyond the centre to teh other side. but due to increasing pull on you, you will turn back and continue to oscillate about the center of the earth.\n\nits good for theory but humanly imposible
1004785
m1*a1+m2*a2=(m1+m2)a\n    47*9.8+6*10^24*x=(6*10^24+47)*x, because final  acceleration after landing does not change for the mass of earth is enormously big. \nThen, 47x= 47*9.8 and x=9.8. Therefore, the acceleration of earth towards the student is 9.8m/s^2
985616
All are true. For the last one,\nsuppose acceleration = a.\nThen the velocity is at + c\nand the position is at^2/2 + ct + d,\nwhere c and d are constants.
999791
Measurements. The acceleration of gravity is:\ng = G*m/R^2\nWhere G is the gravitational constant 6.67*10^-11 N*m^2/s^2, m is the mass of the planet, and R is the distance between yourself and the planet. For the surface of the Earth, m and R are approximately:\nMe = 5.97*10^24 kg\nRe = 6.38*10^6 m\nUsing this, you should get about 9.8 m/s^2.
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
671105
almost exactly, assuming someone on earth is doing the measurement. someone moving very fast would get a different result.
324486
Add v2/R.\nv=speed of rotation at the suface of the earth.\n(it depends on the geographic altitude fi)\nR = radius of the earth.
393015
If a body is fixed at its center of gravity, it is in a state of neutral equilibrium.\n\nTherefore, the whole weight of the body can be assumed to act at the center of gravity.\n\nThe following method is used to find an expression for the C.G of a solid body.\n\nThe solid body is considered to be broken up into elementary particles.\n\nIf m1, m2, m3 etc are the masses of the particles, their weights are m1g,m2g, m3g etc.\n\nThe weights are acting parallel to each other in the downward direction.\n\nIf we assume that the C.G is at the point X,  and x1,x2,x3 etc are the distances of m1,m2,m3, then the  sum of MOMENTS of all these weights about the origin must be equal to the moment of the whole weight of the body about its C.G.\n\nTherefore,   Mg .X =  ( m1g .x1 + m2g .x2+ m3g .x3 + m4g .x4+ ....) where M = sum of all masses.\n\nIn short, the moments of weights of small particles in a solid body about any point is equal to the moment of the whole weight of the body about its C.G.
235638
Remember the formula for centripetal acceleration,\na_c = v^2/r\n\nso if r is doubled but v remains constant, than the centripetal acceleration is halfed.\n\nTo prove this, lets put in some numbers.\nSay v = 2 m/s, r = 1 meter, thus, a_c = 4 m/s^2.\nNow double r to be 2 meters, the new a_c = 2 m/s^2, or 1/2 the original value.
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.
680534
Use the equation from the link below.  You will find that \nv/c = 0.14365, or\nv = 0.14365 c\nor\nv = 43,100 km/s\n\nThe object is moving away from the observer at this rate.\n\nBy the way, you probably won't get credit unless you can work out the math steps I left out.
408862
the force between the object and the earth is dependant on mass, but the acceleration of the object towards the earth is 9.81 meters&#92;second. (ignoring air resistance)\n\nforce = mass x acceleration
931404
Gravity is proportional to the mass of the planet.  I'd say the acceleration on this planet is 0.1045 * 9.81 m/s^2.  \n\nFor b) use x =  1/2at^2\n\nt = time  \na = acceleration due to local gravity
956084
Consider mass m on planet Y\nas also on Earth. We have\nm*g[e]=G*m*M/R^2.........1\nm*g[y]=G*m*[2M]/[2R]^2....2\nDivide eqn.2 by eqn. 1 we have\ng[y]/g[e]\n=2M/M*[1/4R^2]/[1/R^2]\n=2/4\n=1/2
964035
You need to consider vectors ....\n\nAcceleration has a direction. In the case of gravity its down towards the ground. \n\nNow what accelerationis doing is changing the instantaneous velocity of the particle or object or whatever.\n\nVelocity is also a vector.\n\nSo if the velocity is in the opposite direction to the acceleration then the object will motion will slow down and eventually increase in the same direction as the acceleration.\n\nIf the velocity is in the same direction then it will increase in that direction.\n\n So if UP is defined as positive then down is defined as negative. This goes for:\n- position (above (+) and below (-) ground if ground level is defined as zero);\n- velocity is positive if the object is moving up and negaitve oif it is moving downward; and\n- acceleration is positive if applied upward or negative if applied downward.\n\nNote acceleration is in the same direction as an applied force as by Newton's Second Law, F = ma and mass is a scalar quantity .. it has no directionality. It is the same wherever the object is. (Not to be confused with weight which is a force attracting us to the ground and is dependent on the strength of gravity. So for example we weigh on earth and are weightless in space and are 1/6 of our weight on the moon etc ... In all locations our mass is the same but the forces acting on us are different)
58016
The thrust is irrelevant. This is a conservation of angular momentum problem so all you need is the momentum of the planes. When calculating the angular momentum of the earth, you need to multiply by a factor of 2/5 to make up for the fact that the earth is solid and not all of the mass is at the surface.\n\nyou can do the math yourself.