Is the earth's mass constant?
83889
No.  Throughout the years, it gain a bit of mass due to being bombarded by dust and meteorites from space.  This adds up to be several tons on a yearly bassis, but is hardly noticable given the size of the Earth.  It will lose a bit of mass on ocations when we launch a rocket out into space, but that is hardly anything.
798707
Interesting question samsam.\n\nMost people's immediate reaction would be to say that "of course it isn't".\n\nBut I can tell you that the earth increases in mass by 2 tons every single day !\n\nIt is estimated that around 20 tons of space debris in the form of dust, micro-meteorites, ice, and sometimes larger particles, hits the earth's atmosphere very day. Most of this burns up, but an estimated 2 tons of it will find its way to the ground.\n\nI will leave the physicists and mathematicians amongst you to calculate if the resultant changes in gravity will alter our orbit enough to destroy the Earth before the Sun explodes into a red giant !
406753
The average radius of the Earth is about 6378.1 kilometers, or abnout 6378100 meters.\nCompared to this, a few hundred meters doesnt make a significant difference.\n\nFor reference, the mass of the Earth is about 5.9742 E24 kilograms.\nG = 6.67 E-11\n\nThe gravitational acceleration on the surface of Earth (r = r_E = 6378100 m) is,\ng = G*M / r^2 = about 9.795 m/s^2\nIs the object is now 1 km above the surface of Earth (1000 meters higher), the gravitational acceleration is about 9.792 m/s^2, just .003 m/s^2 smaller, barely noticible.
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.
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.
406606
Your question is very awkwardly worded. I assume you're talking about mass, not weight. Mass is the amount of matter in something. Weight is force that gravity "pulls" on it. With the same object, mass is always constant. Weight changes if you're on the moon, or in outer space.\n\nAs for its gravity, do you mean the Earth's pull on it, or the gravity field the object projects? If it's the former, then yes. That's how weight is defined. All things fall at the same rate, however. As for the latter, then, yes, the mass of an object dictates how big the gravity field around it is. The Sun projects a huge gravitational well, while a penny projects next to nothing.
46266
When it is said that that the net force is zero, it means yes, there is gravitational forces acting at the cenetr of the earth, but being at the center, with equal amounts of earth mass in every direction, the forces pulling on you at the center would all cancel each other out, resulting in the <perception> of having no forces acting on you, when in reality, gravity is always at work.
70874
Um. No.  The Earth's core is the centre of the Earth.
701530
Plug the numbers into the Newtons Equation :\n\nF =  (GMm) / r^2\n\nF = Gravitational Force / weight  \nG = Universal gravitational constant   6.673*10^-11 (note the - sign on the exponent)\nM = mass of earth (kg)                 5.98*10^24 kg\nm = mass of satellite (kg)             150kg\nR = radius of earth    (m)             6,378,137 m\nh = height above earth (m)             12,800,000m\nr   = h+R\n\nThe answer will come out as the weight being @ 162.67N\n\nNot weightless, although someone standing on the satellite would feel weightless as its in continuous free fall.\n\nEver been in an elevator and felt lighter as it falls or been weighless on the big dipper as it plummets?\n\nThe gravitational force (weight) of 162.67N is what keeps it in orbit and prevents it from flying off at a tangent. Basically it pulls the satellite towards the earth but the satellite is travelling so fast it misses and continues round in its orbit.\n\nAdded:\n\nWeight is measured in Newtons and is the attractive force between the two masses. (M & m)\n\nIn this case the earth and the satellite. A 1 Kg mass on the surface of the earth will be pulled towards the earth's centre with a force of 9.8N, therefore its weight at ground level is 9.8N. If you plug take m as 1kg, h as 0 and plug them into Newton's formula should get the answer 9.8N, which is where we get the constant g from. \n\nAs you get further away from the centre of the mass the gravity field decreases in inverse proportion to the distance squared. Hence the divide by r^2 on Newtons fornula. Therefore the weight of a 1Kg mass at 12,800km is 1.08N.\n\nBecause the satellite is in a stable orbit the opposing forces of the satellite trying to escape orbit and the gravitational pull towards the earth are equal so yes the weight is 162.7N and the centrefugal force is 162.7N. This is the correct answer to the problem. \n\nAs an aside; for simplicicities case in this scenario the mass of the satellite can be regarded as constant due to its relatively low velocity compared to the speed of light. \n\nHowever as you come closer to the speed of light mass increases according to Einstein's theory of relativity, eventually reaching infinity at light speed.
441441
Be careful not to confuse the concept of weight with the concept of mass. The Earth's weight would depend on the acceleration due to gravity (Here on Earth we call it 'little g' and it's 9.8 m/s squared) on the enormous planet with a huge bathroom scale that you could set the whole planet Earth on. The Earth's mass doesn't change (except for increasing steadily due to cosmic dust accumulation) when you move it from gravitational field to gravitational field.  Likewise your weight changes when you walk on the moon, but your mass stays the same. It's the different acceleration due to gravity  (The moon's instead of the Earth's) that changes your weight.
353194
Well, there are fewer than 10 billion people and the average mass is less than 100kg, so the total mass of people will be less than 1 trillion kg. This is 10^12 kg. The mass of the earth is about 6*10^24 kg, so the total mass of people is less than one 6 trillionth of the mass of the earth. Not a huge effect.
606827
The mass of the earth is 5,974,200,000,000,000,000,000,000 kg\nQuite heavy really.
1117701
the earth does not weigh. It has a mass.\n5.9742×10^24 kg
924812
The mass of the sun is 330,000 times that of the earth. It is about half again as dense as water.
747527
HOW IS THE MASS OF THE EARTH DETERMINED?\n\nNewton, Galileo, Henry Cavendish, and Eratosthenes contributed to this amazing calculation. This calculation is done using Newton's Law of Gravity, which formulates the attractive force (gravitational force) that two masses exert on each other: \nF=GmM/r2 \nIn Newton's equation, F is the gravitational force, G is a constant of proportionality, M and m are the two masses exerting the forces, and r is the distance between the two objects. \nG was calculated by Henry Cavendish in 1798, and was determined to be 6.67 x 10-11 m3/(kg sec2). \n\nAlso needed is Newton's second law of motion, F=ma, where F is the force applied to an object, m is the mass of the object, and a is its acceleration due to the force. \n\nGalileo determined that the acceleration due to the force of gravity of Earth was a constant equal to 9.8 m/sec2 near the surface of the Earth. \n\nLastly, you need to know the radius of the Earth; this was first calculated by the Greek Eratosthenes thousands of years ago (by comparing shadows in wells during the summer solstice about 230 B.C.). \n\n\nCALCULATING THE MASS OF THE EARTH\n1. F = GmM/r2 = ma, where F is the gravitational force, G is the gravitational constant, M is the mass of the Earth, r is the radius of the Earth, and m is the mass of another object (near the surface of the Earth). \n\n2. GM/r2= a (The m's canceled out.) Now solve for M, the mass of the Earth. \n\n3. M = ar2/G, where a=9.8m/sec2, r=6.4 x 10 6m, and G=6.67 x 10-11m3/(kg sec2). \n\n4. M = 9.8 x (6.7 x 106)2/6.7 x 10-11 = 6.0 x 1024 kg \n\nTHE PEOPLE BEHIND THE CALCULATION\nCAVENDISH, HENRY\nHenry Cavendish (1731-1810) was an English chemist and physicist. Cavendish discovered that hydrogen gas was a substance different from ordinary air (whose components he analyzed), described the composition of water (hydrogen and oxygen) and other important works. Cavendish was the first person to determine Newton's gravitational constant gravitational constant and accurately measured of the Earth's mass and density. \n\nERATOSTHENES\nEratosthenes (276-194 BC) was a Greek scholar who was the first person to determine the circumference of the Earth. He compared the midsummer's noon shadow in deep wells in Syene (now Aswan on the Nile in Egypt) and Alexandria. He properly assumed that the Sun's rays are virtually parallel (since the Sun is so far away ). Knowing the distance between the two locations, he calculated the circumference of the Earth to be 250,000 stadia. Exactly how long a stasia is is unknown, so his accuracy is uncertain. He also accurately measured the tilt of the Earth's axis and the distance to the sun and moon, and devised a method for determining the prime numbers up to a given number (the Sieve of Eratosthenes). Eratosthenes made numerous contributions to the sciences and arts in many fields, including geography, mathematics, astronomy, chronography (calendars), music, and literature. Eratosthenes was a brilliant all-around scholar; although not the top expert on any topic, he was well-versed in all subjects, and therefore nicknamed "Beta" (which is the second letter of the Greek alphabet). \n\nGALILEI, GALILEO\nGalileo Galilei (1564-1642) was an Italian mathematician, astronomer, and physicist. He was the first person to use a telescope to observe the skies (in 1609). Galileo discovered four moons of Jupiter, observed the phases of Venus, studied sunspots, and discovered many other important phenomena. After publishing the many discoveries he made using his telescope, including the motion of the Earth around the Sun (the Copernican System), Galileo was accused of heresy by the Inquisition (in 1633). \n\n\nNEWTON, ISAAC\nIsaac Newton (1642-1727) was an English mathematician and physicist who invented calculus (simultaneously, but independently of Leibniz), formulated the laws of gravitation and mechanics, investigated the nature of light (he discovere
695350
Yes.\n\nIn fact there is a very nice relationship that when you are below the surface of the Earth (assuming the Earth is a perfect sphere), your weight is proportional to your radius from the center. E.g. if you were 1/3 of the way to the center (2/3 from the center), you'd weigh 2/3 as much as you do on the surface.\n\nThe proof uses a variant of Gauss's law from electromagnetics (because Newton's equation for gravity and Coulomb's equation for charge have the same form). Without delving into advanced calculus, the gravity that affects you depends only on the earth in the sphere below you - everything at your altitude and above cancels out. Since the volume of a sphere is proportional to r^3, but the gravitational force is proportional to 1/r^2, the remaining term is simply r - so your weight is proportional to the radius, if you're inside a sphere.\n\nOf course this assumes that the Earth's weight is distributed evenly. However, it's close enough to even for most purposes.
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