What kind of forces are present on the Earth, Moon and Mercury?
827409
We know of four types of fources, gravitational, electromagnetic, strong and weak forces. gravitational force will be there whereever massive bodies are present. If electrically charged bodies are present then electromagnetic force will also be there. Strong nuclear and weak forces are also likely to be there between subnuclear particles. So I guess all four forces will be present at the planets you mention.
748344
Every particle has a gravitational force associated with itself. It the basic nature of the particle.\nThe force is given by : F = G(m1*m2)/r^2\nwhere G= Universal constant of Gravity.\nSince the earth is very huge, the force between 2 humans is negligible as compared to the each of them to earth
1175736
the gravitational force can act in any medium as well as in vacuum. the sun exerts a gravitational force on the earth across a vast expanse of vacuum! this itself is proof enough.\nthe objects will all touch the ground at the same time whether in a medium or in vacuum. this occurs for the sole reason tht all of them are of equal mass.
878342
Nuclear force, gravitational force, (I think these can be either attractive or repulsive:) electrostatic force, and magnetic force.
578217
there is no place on this planet, or any other, that is not subject to tidal forces; whereever bodies are in orbit then gravity will produce tides.\n\nTides affect the land as well as any body of water, the effect may not be as obvious, but it is there.
1053094
What you're describing is called the 'grand unified theory,' or GUT. There are four known fundamental forces in nature that literally control everything--gravity, electromagnetism, the weak nuclear force, and the strong nuclear force. Science has been successful in uniting all of these except gravity.
39089
75 newtons (around 17 pounds)....the moon's gravitational pull is only about 1/6 that of earth.
1097470
All Mass generates a gravitational pull.\n\nBecause the Moon is so close to the Earth, we come into range of its pull.\n\nWith the moon in pulling in a different direction to Earth's pull, there is a minute difference which is discernible in water because it is continually moving and has a very small mass in itself.\n\nSo when the Moon is above us, the gravitational pull of Earth is slightly less to that the tide level will increase.
180202
Neither. It's the mass of the planet itself. Everything that has mass projects a gravitational field. The more massive an object is, the bigger its gravitational "pull". This is why objects like the Sun can support several large planets, while Pluto (which has a moon about a third the size of it) gets pulled around by its moon.
829728
As far as I am aware, there are none in the terrestrial sense, after the huge meteor impact that formed the Caloris Basin there were massive extrusions of flood lavas that formed the Smooth Plains...but these were 'seeps'.  The core of Mercury is thought to have only a small amount of magma which gives the planet it's magnetic field...but it is doubted that without the outside intervention of the major meteor strike that there would have been any lava flow at all.\n\nIt is also thought that seismic vibrations from the Caloris impact are what caused the hilly and furrowed terrain on the opposite side of the planet.\n\nIt is remarkably similar to our own moon in many respects, airless, crater impacted and causing tides on the sun, just as the moon causes tides in our oceans.
1102968
The tidal forces of both the Moon AND the Sun.
30388
Gravity, the strong nuclear force, the weak nuclear force, and electromagnetic force.
1027284
because the sun's gravitational pull is much greater than the earth's...same reason the moon goes around the earth and not the other way around
583414
There is an influence, but it is difficult to detect.  First of all, I'll point out that Foucault's pendulum does not relate to this question.\n\nSo, what is the force?  The Earth moves under both the Sun's and Moon's gravitational influences, but it's the centre of mass of the Earth which falls freely through the gravity fields.  An object on the surface of the Earth experiences a tidal force because it's, generally, not at exactly the same distance from the Sun or Moon as is the centre of the Earth.\n\nLet's consider the Sun, the Moon's case is exactly analogous.  If the pendulum is directly between the Sun and the centre of the Earth, it feels its maximum tidal force, 5.1e-7 Newtons per kilogram.  Unfortunately, that force is vertical, along the pendulum's suspension string/rod, and so is unmeasurable by deflection.  Well, what if the pendulum were 90 degrees from the Earth-Sun line.  Then the force is transverse, and that's best for measuring the force, but because the pendulum is now basically at the same distance from the Sun as the Earth, the tidal force is essentially zero, which doesn't help at all.\n\nSo, we want to calculate the transverse force on the pendulum due to the Sun, as a function of the angle between the pendulum-Earth and Sun-Earth lines.  This is\n\n2 * G * M_sun / R^2 * (r sin T cos T ) / R\nWhere R is the distance from the Earth to the Sun and r is the radius of the Earth.\n\nMaximize this function with respect to the angle T, and we find that the optimal angle is at 45 degrees.  Then, the transverse force is 2.5e-7 N/kg.  Your apparatus has to be capable of detecting this transverse force in the context of a vertical force of 9.81 N/kg.  Your deflection angle will be 2.6e-8 radians.  If you have to measure a deflection of 1 millimetre to see the effect, you will need a pendulum arm some 39 kilometres long.\n\nThe force of the Moon is roughly double this, but you'd still need a tower 20 kilometres high from which to suspend your pendulum.\n\nWith sufficiently sensitive instruments, the tidal force of the Moon can be measured.  In fact, the Moon was "discovered" by the folks at CERN, because of Earth tides.  See http://bulletin.cern.ch/9809/art1/Text_E.html
140176
The Sun, Earth, and Moon are all gravitationally tied together. In other words, the interaction of their individual gravitational fields closely relates one to the other. The Earth/Moon system moves around the Sun in a particular orbit. Should the Sun somehow disappear, the movement of the Earth/Moon system would be radically altered. Even the Sun, the largest mass of the three objects, has its movement through space influenced by the gravity of the Earth and Moon. If we could view our Sun from several light years away, we'd notice that it moves through space not in a straight line, but instead a wavy line as the Earth/Moon system tugs at it.
86134
Potential is measure by mass x gravity x height. Since the force of gravity on the moon is about 1/6 of the earth, the potential on the moon compared to that of the earth would be: mass x (1/6)gravity x height.
894961
The gravitational force (Fg) exerted by Saturn on its moon is\n\nFg = G*M*m/r^2\n\nwhere G is the universal gravitation constant, M is the mass of Saturn, m is the mass of the moon, and r is the radius of the orbit.\n\nThe gravitational force is balanced by the centripetal force (Fc) due to the motion of the moon in a 'circular' orbit around Saturn.\n\nFc = m*v^2/r \n\nwhere v is the orbital speed, which can be expressed as\n\nv = 2*pi*r/T\n\nwhere T is the period of the orbit. Inserting this into the equation for the centripetal force you have\n\nFc = 4*pi^2*m*r/T^2\n\nNow since\n\nFg = G*M*m/r^2 = 4*pi^2*m*r/T^2 = Fc\n\nThe mass of the moon, m, cancels leaving\n\nG*M/r^2 = 4*pi^2*r/T^2\n\nwhich can be rearranged to give Kepler's third law that says that the square of the orbital period is proportional to the cube of the orbit's semimajor axis (taken as the radius for this problem, which assumes a circular orbit). Solving for the mass of Saturn\n\nM = 4*pi^2*r^3/(G*T^2).