stone at the earth’s center compare with their values at the earth’s surface?
2006-08-10 08:18:21 · 8 answers · asked by Twhat 1 in Science & Mathematics ➔ Physics
The mass of stone remains unchanged.
Assuming that the earth is perfectly uniform and spherical the weight of the stone would be zero. However, since this isn't the case, the stone would still have some weight in a real life experiment.
Note to wdmc: In the earth's case, weight wouldn't decrease linearly since the earth's core is denser than its crust and mantle.
2006-08-10 09:37:25 · answer #1 · answered by bromothymol 4 · 1⤊ 0⤋
Responder wdmc is closest to being right. The weight does indeed reach zero at the center, but the decrease is not exactly linear with depth because the earth is not of uniform density. The gravitational pull felt by an object at any depth is the same as if all the material above the object (that is, farther from the center of the earth than the object is) were not there.
2006-08-10 09:36:40 · answer #2 · answered by Anonymous · 0⤊ 1⤋
Can't be done, the depth to the centre of the earth is some 6400km, the deepest hole drilled into the earth reached a depth of some 12 odd km... at these depths the pressures and temperatures are way more than current technology can overcome...
Oil & gas wells are quite regularly drilled to depths of 4000 - 5000m and the never ending quest for more oil spurred by the high oil price is driving exploration deeper but getting to 6000m and beyond is a challenge...
2006-08-10 09:52:16 · answer #3 · answered by Farlig 2 · 0⤊ 2⤋
humorous how such diverse people imagine they comprehend the answer to this yet do not. The stone will hit the fringe of the tunnel. It begins with angular momentum from being dropped by technique of someone on the floor of a rotating Earth. That angular momentum is conserved. because the stone receives in route of the midsection of the Earth, a similar angular momentum will correspond to a higher angular %, equivalent to a spinning skater pulling their hands in. The stone will overtake the east fringe of the tunnel. The result is equivalent to the Coriolis result. it ought to leap off. it may leap backward and forward some cases. The ability lost in friction and bouncing will avert the stone from having adequate ability to emerge from the alternative fringe of the tunnel. it ought to oscillate some cases around the midsection of the Earth, yet will ultimately settle on the midsection.
2016-10-15 11:55:30 · answer #4 · answered by ? 4 · 0⤊ 0⤋
The mass of the stone stays the same.
The weight of the stone decreases linearly as it falls until it reaches 0 newtons at the center of the earth.
2006-08-10 08:30:13 · answer #5 · answered by wdmc 4 · 0⤊ 1⤋
mass is a function of the reaction of matter with the higgs field (asociated with the higgs boson) it is not a function of location. no change in mass
weight is the attraction a mass feels. at the center of the earth, asuming a round earth, the attractions cancle and the stone would be weightless
2006-08-10 08:33:28 · answer #6 · answered by robert m 2 · 0⤊ 0⤋
The mass is unchanged.
The weight decreases to zero as it approaches the center of earths gravitational field --- if that is the "center of the earth", then the answer is zero.
2006-08-10 08:29:32 · answer #7 · answered by dbear 2 · 2⤊ 0⤋
what i think is that mass never changes whereas weight does.
bec weiht is mass multiplied by gravity and gravity changes with change in depth n height.
2006-08-10 08:31:13 · answer #8 · answered by ms 1 · 0⤊ 0⤋