If we consider earth mass quasi constant  What is total its total KE including spin energy?
339126
KE = 1/2 mass* v²\n\nThere are two velocities.  One is spin energy, however this is a tricky one since velocity is relative to radius and polar angle.  The other is the energy of the motion around the sun, which is somewhat easier to calculate.  The earth's velocity is 30 km/s, so:\n\nKE = 1/2M* (30 km/s)² = 450M (km/s)²\n\nFor spin energy:  KE = 1/2Iω²\n\nWhere I is moment of inertia and ω is angular velocity.\n\n"As an example, let us calculate the rotational kinetic energy of the Earth. As the Earth has a period of about 23.93 hours, it has an angular velocity of 7.29×10-5 rad·s-1. Assuming that the Earth is perfectly spherical and uniform in mass density, it has a moment of inertia, I = 9.72×1037 kg·m2. Therefore, it has a rotational kinetic energy of 2.58×1029 J."
362695
The solar constant for Earth is sc = 1360 W / m2 = 1.36 GW / km2 [1]. Assuming that the radius of the Earth is RE = 6,366 km then the total rate of energy flow (power) on Earth is equal to: \n\n\nPE = sc    RE2  1.7  108 GW (1) \n\nSome of this energy accumulates in the Biosphere, but most of it is radiated back to space in the form of electromagnetic radiation. \n\nThe temperature of the Sun's surface is TS  5800 K. Assuming that the Sun is a black body at temperature TS = 5800 K and that all of the energy impinging on Earth from Sun is radiated back into space in the form of black body radiation it can be shown that the Earth black body temperature TE is [1] : \n\n\nTE = (0.25) 1 / 4  (RS / RSE) 1 / 2  TS  280 K (or  7oC) (2) \n\nwhere: RS  7  108 m - radius of the Sun, RSE  1.5  1011 m - average Sun-Earth distance \n\nThe Sun is the major energy source for Earth. Geothermal energy is comparatively minor, since ice caps formed on the Earth's Poles where the average intensity of the Sun's radiation is low.
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.
986703
In most physics homework problems, energy is conserved, and the following is true:\n\nEo = Ef\n\nHowever, in this problem, energy is lost due to friction.  Therefore...:\n\nEo - Q = Ef\n\nWe know that the child had potential energy at the top, and kinetic energy at the bottom.  Use this information to calculate the energy lost due to friction.  The energy lost due to friction will be denoted "Q" (Q usually stands for heat, which IS thermal energy):\n\nEo - Q = Ef\n\nRearrange the expression, since we're solving for F:\n\nQ = Eo - Ef\n\nNow plug in the formulas for potential and kinetic energy:\n\nQ = mgh - (1/2)mv²\n\nYou don't have to do this, but I factored out the "m" (mass):\n\nQ = m[gh - (1/2)v²]\n\nYou can plug things in now, since the values are all given:\n\nQ = 19.6[(9.80)(3.80) - (1/2)(2.52)²]\n\nQ = 668 J
962133
E potential = mgh\n= 4 kg ( 9.81 m/s^2) ( 2.5)\n=98.1\n\nPotential energy was not lost....it transformed into Kinetic Energy\n\nKinetic + Potential = Total Energy\n\nLaw of Conservation of Energy
558215
F = KM1M2/r^2\n\nwhere K is the constant, M1 and M2 are the magnetic moments, and r is the distance between them.
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.
437310
Every mass in motion is said to have mechanical energy.\nHowever a mass is also structure ed with atoms wich are also moving with mechanical energy. The is the energy of the structure. You can represent it a the total mass  time velocity of light squard. Usally in the motion of masses the mass changes are very small. Hence you can consider mass as a constant. The mass time the final velocity squared of the moving mass is the mecahnical energy of maas in motion. assuming there is no friction. if there is friction you must substract the energy loss due to friction.
665122
mass of Earth = 5.9742 × 1024 kilograms
17651
I assume here you're asking about the Earth's mass (mass and weight are different concepts and often confused).  The earth's mass is 5.9736 x 10^24 kg.
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
256338
a) when it it closest to the gravitational source\nb) when the potential energy is minimum (same as a)\nc) speed = |velocity|\nd) momentum(p)= mv\ne) KE = (mv^2)/2\nf) usualy the potential energy is defined as 0 at infitity\ng) conservation of energy\nh) when it it closest to the gravitational source\ni) conservation of angular momentum\n\nThat should be enouph to figure out the answers... but think about it and draw some pictures so that you know it is right.
61784
That's an interesting qts. I assume that you wish to invlove pure physics in this qts. Forget all the heat at the core of the earth and also air resistance.\nConsider pure gravitation theory.\n\nIf say the fellow who fell in the hole has mass m\n\nThen force of gravity exerted by the earth is :\n\n  F = GMm/(r*r) ..... Where G - universal gravitation constant\n\n                            M - Mass of the earth\n                            r - distance between earth and body of mass m\n\n          acceleration due to gravity\n            = (1-h/R)g   .... where h is distance of body beneath the earth's surface and g is acceleration due to gravity at the surface of the earth.\n\nSo as the body comes closer to the core of the earth, h tends to R.\nSo the acceleration of the body goes on decreasing very much. So, body may not actually hover, since his velocity will be constant the instant it is at the core.
90476
It is not effected to any significant degree.\n\nThe numbers used to do any calculations would absorb any effect which may occur.  For instance the mass of the Earth is about 5.9742 E24 kg.  But do you really think this is an exact number, no, it is just an estimate rounded off.  To try to use this value for the Earth's mass to try to calculate the effects of something so insignificant will not work too well.\n\nWhen you are standing on the Earth, you and the Earth together have a certain angular momentum.  When you jump, angular momentum must be conserved (its the law).  But now that your mass is rotating slightly farther out than it used to be, your angular momentum changes, there for the Earth's must change ever so slightly to compensate.\n\nIf you wanted "numbers and calculations", you did not specificy nearly enough information.
815939
Not sure I understand your question....\nThe total mass of the earth is 6x10^24 kg.\nThe total human world population is about 6.6 x 10^9. Assuming an average weight of 50 kg, this give a total human mass of 3.3 x 10^11 kg, far smaller than the earth.  \nThe total biomass (all plants and animals together) is estimated here\nhttp://en.wikipedia.org/wiki/Biomass\nat about 2x10^15 kg, still 9 orders of magnitude smaller than the mass of the earth.
464446
Gravitational potential = G *Mass of the earth/radius of the earth.\nIn other words its equal to Gravitational energy per unit mass.
446986
Yes, it does, but by a very negligible amount.\n\nDue to the law of conservation of angular momentum, the Earth has some over all value for its angular momentum (and for that matter, so does the solar system and our entire galaxy, ... as well).  As thr plane is flying around the Earth, it has some angular momentum too, different from its original value just sitting still on the surface.  In order for the overall angular momentum of the Earth to remain constant, the Earth's rotation is effected ever so slightly.\nHowever, the effect is so small, it is not even noticible when we are talking about objects as miniscule as an air plane compared to the Earth.\nAlso, since, over all, on average, for every plane flying one way, there is another plane flying the other way (or when the plane returns), so the effects are all canceled out.\n\nIt is the same principle at work is your on a merry-go-round and you move closer in to the center or farther away (or in this case, around while spinning on the merry go round), angular momentum needs to be conserved, it is the law.