What is the acceleration of all objects tward the moon? And same question for Pluto?
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towards the moon or towards the sun?\n\nand i think its centripetal force. (or however you spell it)\ngood luck.
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pluto's volume is 7.15 x 10^9 km^3\npluto's overall density is about 2 g/cm^3\n\nlook here:\nhttp://en.wikipedia.org/wiki/Pluto\nhttp://en.wikipedia.org/wiki/Pluto#Physical_characteristics
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The distance between Earth and Pluto varies as the two planets move around their orbits. When they are closest, they are about 29 A.U. apart; at most, the distance is about 51 A.U. Since light travels about 300,000 kilometers each second, it would take light anywhere between 8 hours and 14 hours to make the round-trip journey to Pluto from Earth.\n\nRadio signals travel at the same speed as light. If you were using radio signals to control a robot on Pluto, you would have to be very patient. If you sent your robot a command to move, the command would arrive at least 4 hours later. Then, if your robot took a picture and sent it back to Earth, it would take at least 4 more hours before you could see that picture! \n\n\n\nAU\n\nAU, which stands for "astronomical unit", is a unit for measuring distance. One AU is the average distance from the Sun's center to the Earth's center. It is equal to 149,597,871 km (92,955,807 miles).\n\nAUs are often more convenient to use than kilometers when measuring large distances such as those in space. In this case kilometers are just too small - it would be like measuring the distance from Boston to San Francisco in inches. AUs simply make a measurement easier to understand and give you something to compare it to.\n\nFor example, Saturn's orbit around the Sun has an average radius of 9.5 AU, which means that Saturn is about ten times farther from the Sun than Earth is. The average distance from the Sun to distant Pluto is about 40 AU. Mercury, the planet closest to the Sun, orbits at an average distance of 0.39 AU.\n\nAUs are generally used for measurements of distances within our Solar System. Distances to stars are much larger, and are expressed in terms of light years. One light year is equal to more than 63,000 AUs. The nearest star, Proxima Centauri, is just over 4 light years away.\n\nAlthough Pluto was discovered in 1930, limited information on the distant planet delayed a realistic understanding of its characteristics. Today Pluto remains the only planet that has not been visited by a spacecraft, yet an increasing amount of information is unfolding about this peculiar planet. The uniqueness of Pluto's orbit, rotational relationship with its satellite, spin axis, and light variations all give the planet a certain appeal.\n\nPluto is usually farther from the Sun than any of the nine planets; however, due to the eccentricity of its orbit, it is closer than Neptune for 20 years out of its 249 year orbit. Pluto crossed Neptune's orbit January 21, 1979, made its closest approach September 5, 1989, and will remain within the orbit of Neptune until February 11, 1999. This will not occur again until September 2226.\n\nAs Pluto approaches perihelion it reaches its maximum distance from the ecliptic due to its 17-degree inclination. Thus, it is far above or below the plane of Neptune's orbit. Under these conditions, Pluto and Neptune will not collide and do not approach closer than 18 A.U. to one another.\n\nPluto's rotation period is 6.387 days, the same as its satellite Charon. Although it is common for a satellite to travel in a synchronous orbit with its planet, Pluto is the only planet to rotate synchronously with the orbit of its satellite. Thus being tidally locked, Pluto and Charon continuously face each other as they travel through space.\n\nUnlike most planets, but similar to Uranus, Pluto rotates with its poles almost in its orbital plane. Pluto's rotational axis is tipped 122 degrees. When Pluto was first discovered, its relatively bright south polar region was the view seen from the Earth. Pluto appeared to grow dim as our viewpoint gradually shifted from nearly pole-on in 1954 to nearly equator-on in 1973. Pluto's equator is now the view seen from Earth.\n\nDuring the period from 1985 through 1990, Earth was aligned with the orbit of Charon around Pluto such that an eclipse could be observed every Pluto day. This provided opportunity to col
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A) Galaxy, about 100,000 years.\n\nB) Sun, about 5 seconds\n\nC) Solar System, it depends on if you mean the orbit of Pluto or all the way out to the Oort cloud.
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http://www.newton.dep.anl.gov/askasci/ast99/ast99241.htm\n\nAnswer:  The average orbital radius of Pluto is 3660 million miles; however, its \norbit is more eccentric than the other planets, so much so that it is currently\ninside the orbit of Neptune (radius 2793 million miles) and will remain so until \n1999.\nHawley\n\n1KM = 0.6M
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Far less than a light year!\n\nConsidering that Pluto is on average about 5913520000 Km from the Sun and that 1 LY equals 9460800000000 Km, you have that Pluto is 0,000625055 LY far from the Sun.\n\nYes, 1 LY is a huge distance.
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First you want to know the acceleration due to gravity of the planet Glob. At the surface the gravitational acceleration (g) can be determined by equating the weight of an object with the gravitational force law\n\nm*g = G*M*m/R^2\n\nwhere R is the radius of Glob, M is Glob's mass, G is the universal gravitational constant, and m is the mass of the object. Solving for g yields\n\ng = G*M/R^2\n\nNow, throwing an object into the air, and neglecting air resistance, we can consider conservation of energy. The kinetic energy of the object when it is released by the astronaut is completely converted into potential energy when the object reaches the top of its travel.\n\n0.5*m*V^2 = m*g*H\n\nwhere V is the initial speed of the object and H is the height to which it traveled, and g is the gravitational acceleration solved for above. Solving for V we find\n\nV = sqrt(2*g*H).\n\nThe satellite traveling in a circular orbit of radius D above the planet has its centrifugal force balanced by the gravitational force of the planet.\n\nFc = m*v^2/D = G*M*m/D^2 = Fg\n\nSolving for the speed of the satellite (v) needed in order to maintain orbit at radius D yields\n\nv = sqrt(G*M/D).
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Solve for the mass of the planet.  You are given a diameter and are told that it is sphereical, so solve for the solume.\nV = 4/3 pi*r^3\n\nOe you know the volume, using its density, solve for the mass,\nD = m/ v\n\nFind the orbital distance of the moon from the planet's center of mass.\nThis distance, r, would be the moons distance above the planet's surface, plus the planets radius.\n\nFind the grvitational acceleration caused by the planets mass on the moon using Newtons law of gravitation,\ng = G*m/r^2,\nwhere m is the mass of the planet, r is the distance from the center of mass, and G is the Universal gravitational constant (6.67 E-11).  Solving this will give you g, the gravitational accelerating the moon experiences towards the planet.\n\nWe are told the orbit is perfectly circular, so we can solve for the orbital speed of the moon by setting the centripetal acceleration eqaul to the gravitational acceleration.\ng = v^2/r\nwhere v is the orbital speed and r is thr radius.
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Are you taking into account the radius of the moon? The equations you've provided treat "r" as the distance between the centers of mass of each object. Therefore:\n\nr = radius of the moon + distance from it's surface\n\nr = 1785 km + 111 km\n\nr = 1,896,000 m\n\nSubstituting this into your equations provides:\n\nT = 2*pi*sqrt((1896000)^3/(6.674x10^-11 * 7.3x10^22))\n\nT = 7431.6 s = 123.9 min\n\nAdditionally:\n\nV = sqrt((6.674x10^-11 * 7.3x10^22) / 1896000)\n\nV = 1603 m/s\n\nHope that clears things up a bit! :)
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I will state this here, in yet another question: THERE IS NO SUCH THING AS CENTRFUGAL FORCE!! Look it up in a basic physics textbook like Giancoli or Serway. Or even do a google search. You have to understand what reference frame you're talking about. The fact that your body tends to want to move to the outside of the curve when your car goes around in a circle has nothing to do with anything other than the fact that your body wants to keep going along in a straight line (conservation of momentum), but the car gets in front of it at every turn (semi pun intended). What is actually happening is that the car is an accelerating reference frame.\n\nAs for the moon/earth question... imagine a man 6' tall throws a tennis ball horizonatally with respect to the earth's surface. The ball will go forward a certain distance (because of it's velocity tangential to the earth's surface) as it falls the ~6' to the ground (because gravity pulls it down). The faster he throws the ball (horizontally), the further it will go before it hits the ground. If he throws it fast enough then the amount that the ball travels forward and the amount that it drops towards the earth's surface can match the curvature of the earth, and the ball will orbit the planet (neglecting air resistance, as we physicists always do) and come around and bonk the thrower in the back of the head. The moon is just a much larger tennis ball, thrown by a much larger man.\n\nYour eqn. is correct, but it only describes the force acting upon the moon. It is not the equation of motion, which detirmines what actually happens to the moon. The equation of motion is:\n\nF = m * a  (F = force, m = mass of moon, a = acceleration of moon around earth)\n\nIn the case of the circular motion of the moon around the earth, "a" is defined by:\n\na = (v^2)/r (where v is the magnitude of the velocity of the moon and r is the center-to-center distance from the earth to the moon)\n\nPlugging one into the other you get:\n\nG * M * m/r^2 = m * (v^2)/r (where M is the earth's mass)\n--> G * M/r = v^2\n\nSo, as long as the tangential velocity, v = sqrt(G * M/r), the moon falls just enough towards the earth as it travels just enough tangentially second by second, that it continues to orbit around the earth. (Note: there is NO CENTRIFUGAL FORCE anywhere in any of these eqns.!!)
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The Rate of change of momentum of a body is equal to the resultant force acting on the body, and takes place in the direction of the force.\n\nMomentum = mass * velocity\n\nThe masses are the same on Earth and the moon.\n\nMomentum would be equal!
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Compared to the stars the moon moves from west to east.\nIn 27.3 days it returns to it 'start star'.\nIn one night it moves (360 / 27.3) degrees.\nIn one night it moves 13,2 degrees from west to east compared to the stars.
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Highest on Pluto and lowest on Jupiter. \n\nOK, Pluto isn't a planet any more, but you did say bodies and not planets. OK, Jupiter has no solid surface, at least not until thousands of miles down in the planet. OK, on a small asteroid you could jump higher than on Pluto. In fact,on a small enough asteroid you could jump off it and never fall back because you have exceeded escape velocity. So highest would really be from any of millions of asteroids or meteoroids. But I digress.
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That greatly depends on the kind of spacecraft you use. The Apollo missions took about three days to reach the moon, but the New Horizons spacecraft (the one sent early 2006 to study Pluto) is the fastest man-made object ever made, it went past the moon's orbit in merely nine hours.
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Pluto is 6 Billion kilometers away from the sun.  Light takes about 5.5 hours to reach Pluto.  Which means Pluto is only 6 light HOURS away from the sun.
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33.3 lbs\n\nThe moon's gravity is 1/6 that of earth.
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It would have to be a VERY large rock to have an acceleration of 2 m/s^2. That is even greater than the moon's gravitational acceleration. Even if your rock was incredibly dense, it would still have to be huge. I think this has to fall in the realm of pure fantasy.