The value of g at the Earth's surface is about 9.8N/kg^2...?
914835
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.
710717
GPE is the gravitional potential energy.\nM is the mass.\nG is the gravitational constant.\nH is the height.\nJ stands for joules.\n\nGPE=(500 + 85 kg)(9.8 s^2)(1000)\nGPE=5733000 J
965597
you convert mass into force by multiplying with 9.8m/sec^2 which is the acceleration due to gravity
1871
To put it plainly and simply quoted:  \n\nF = Gm1m2/r2 = ma\n\nGm/r2 = g\n\nm = gr2/G\n\nm = (9.8 m/s2)(6.37 x 106 m)2/(6.673 x 10-11 Nm2/kg2) \n\nm = 5.96 x 1024 kg\n\nThe Earth gains mass each day, as a result of incoming debris from space. This occurs in the forms of "falling stars", or meteors, on a dark night. The actual amount of added material depends on each study, though it is estimated that 10 to the 8th power kilograms of in-falling matter accumulates every day. The seemingly large amount, however, is insignificant to the Earth's total mass. The Earth adds an estimated one quadrillionth of one percent to its weight each day.
987931
you are making a mistake with the units\n\ndid you mean kg/s^2? if it is so the answer is 10300 kg/s^2\nif did you mean kgf/m it is 1050 kgf/m \nif not your units are wrong. \nremember that 1 N (newton) is 1 kgxm/s^2\nand 1 kgf = 9.8065 kgxm/s
73747
if the weight of the person is measured on Earth to be equal to 85 N then the mass is to be equal to \n\nm = F/a = 85/9.8 = 8.67 Kg\n\nif the weighting is taken on the moon then the mass is equal to \n\nm = 85 / 1.67 = 50.9 Kg
1117701
the earth does not weigh. It has a mass.\n5.9742×10^24 kg
1061210
If you're not taking earth rotation into account, the value of GMe is 9.8 * (6.38E6)^2, so the square of the radius required is that value divided by .660, the root of which is 2.4584521E7.  Subtracting the earth radius from that gives\n\n1.8204521E7, or about 18 million km.\n\nIf you're considering a point that stays directly above the earth as it rotates, the nswer depends on the latitude and is the solution of a cubic equation.   I'm not going there for you!
784060
Mass (kg) 5.688e+26 \n\nAsking or saying the mass of a planet has more meaning.  You can't really talk about the weight of a planet.
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.
864950
Gravitational Potential Energy = m*g*h joules\n        where   m - mass of the object in kg\n                     g - accelaration due to gravity in m/s^2\n                     h - height of the object from the reference in m\nHere the height from reference is 500m\ng is taken to be a constant and its value is 9.81m/s^2 though it may vary slightly with latitude, altitude and other such factors.\nMass is given as 0.400kg\nThus Potential energy = 0.400 * 9.81 * 500\n          Potential Energy = 1962 joules\n\nHope so u got wat u needed.
888331
The previous suggestion of calculating the potential energy is good - I started doing it that way but realized this was easier:\n\nLet x be the distance along the tunnel from its center. The acceleration of gravity anywhere in the earth is gr/R pointing radially downward where r is the distance to the center, R is the earth radius and g the acceleration of gravity at the surface.\n\nTo get the component of this force along the tunnel it has to be multiplied by the cosine of the angle of the tunnel with the straight downward (radial) direction which is just x/r. In other words, the acceleration is just:\n\na = -gx/R\n\nThis has the same form as the force for a simple harmonic oscillator with spring constant k=g/R and unit mass and that is known to have period:\n\nT = 2pi/sqrt(k) = 2pi sqrt(R/g)\n\nThe trip from one endpoint to the other takes one half a period so this time is:\n\nt = pi sqrt(R/g)\n\nR is about 6400km and g = 9.8 m/sec so this seems to give about 42 minutes. Amazingly enough it seems to be the same for any straight line path connecting two points on the surface.
881319
force=mass x acceleration(a)\nweight=mass x acceleration due to gravity(g)\n\nweight = force\nSo unit of weight is newton(N).\n\nValue of acceleration due to gravity(g) on the earth is 9.8 m/s^2\n\nValue of acceleration due to gravity(g) on the moon is 1/6th value acceleration due to gravity(g) on the earth and that is 9.8/6=1.6 m/s^2\n\nMass of a material remains same every where.\nWeight of meteorite on moon is 65/6=10.83N
377356
Surface gravity of sun: 273.95 m s-2 (27.9 g)
366327
To convert between g and kg you just need to know that 1 kg = 1000 g.  If you know your metric conversions, these become very easy!  One "kilo" of anything is 1000 of that base unit.  So 1000m = 1 km as well.  So to find out how many grams 8.2 kg is just think: \n\n8.2 kg X (1000 g/ 1 kg) = 8,200 g
101436
The mass of the Earth is roughly 5.9742 E24 kilograms.\n\nWhy the need for such an accurate number?
816057
again ... conservation of (mechanical only here) energy\n\npot energy = kin energy\n\nmgh = 1/2 m V^2\n\nso V = sqrt(2gh) -- see mass irrelevant\n\nV = sqrt(2*9.8*8.53) = 12.9 m/s\n\ng=9.8m/s^2 is the acceleration due to gravity\n\ng=10m/s^2 is not a bad approximation