What would the astronaut weigh on a world which is 3 times as massive as the earth and twice its size..and why
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.
926787
Mass is a constant.  \n\nWeight depends on gravity.\n\nSo, your mass is the same on earth and on the moon,\nbut you would weigh one-sixth as much on the moon\nas you do here on earth.
681101
5.98 sextillion metric tons\n\nEarth's Weight (Mass): 5.972 sextillion (1,000 trillion) metric tons. That's 5,972,000,000,000,000,000,000 tons! Actually, scientists prefer to refer to this measurement as the Earth's mass instead of weight since weight is the result of Earth's gravitational pull on another object. And the Earth cannot pull on itself! As the Earth orbits the Sun, it is weightless. If the Earth were placed on the Sun, it would weigh more than if it were placed on Jupiter, the largest planet in the solar system but much smaller than the sun. Yet, Earth (or any other object for that matter) would have the same mass regardless of where it is located.\n\nAge: About 4.5 Billion Years Old\n\nDistance from the Sun (average): About 93.1 million miles (also one Astronomical Unit or AU). It is the third planet from the Sun.\n\nFarthest Distance from the Sun: 94.5 million miles.\n\nClosest Distance to the Sun: 91.4 million miles. \n\nSpeed through Space (around the sun): 18.4 miles per second or about 67,000 miles per hour.\n\nSolar Orbit: It takes Earth 365.2422 days to orbit the sun. This is the basis for the year. \n\nRotational Speed: About 1,070 miles per hour at the equator.\n\nRotational Time: It takes the Earth 23 hours, 56 minutes and four (4) seconds to make one complete 360° rotation.\n\nRotational Tilt: 23.5° on its axis, a straight line through the planet from the North Pole to the South Pole. The tilt is in relation to Earth's near circular orbit around the sun.\n\nEarth's Size, Distances & Surface Features\n\nEquatorial Diameter: 7,928 miles\n\nPolar Diameter: 7,901 miles\n\nCircumference: 24,907 miles\n\nSurface Area: 197 million square miles\n\nHighest Point: 29,028 feet above sea level, Mount Everest, formed 60 million years ago, located on the border of Tibet and Nepal in the Central Himalayas in southeast Asia.\n\nLowest Point (on Land): 1,320 feet below sea level, Dead Sea, located on the border between Israel and the West Bank to the west and Jordan to the east. It is so salty -- the saltiest on Earth -- that it is unable to support any type of life.\n\nDeepest Point on Earth: 35,802 feet, Mariana Trench in the Pacific Ocean. The water pressure there is over eight tons per square inch.\n\nEarth's Vital Systems & Elements\n\nSurface Data: 70% covered by water, with the remaining 30% composed of the seven continental land masses.\n\nWater Composition: 97% salt water, 3% fresh water (only .3% of all water is usable by humans).\nTotal Water Supply: About 326 million cubic miles. \n\n~ Oceans: 317 million cubic miles\n\n~ Ice Caps, glaciers: Seven (7) million cubic miles\n\n~ Ground water: Two (2) million cubic miles.\n\n~ Fresh Water Lakes: 30,000 cubic miles\n\n~ Inland Seas: 25,000 cubic miles\n\n~ Soil Moisture: 16,000 cubic miles\n\n~ Atmosphere: 3,100 cubic miles\n\n~ Rivers: 300 cubic miles (Almost all of humans' drinking water comes from rivers.)\n\nAir Composition: 78% nitrogen, 21% oxygen, 1% other content (including greenhouse gases). The element Argon makes up about 93% of the other content, with carbon dioxide - the primary greenhouse gas, accounting for a little over 3%. Argon is an odorless and inert (inactive) gas that is commonly used in incandescent and fluorescent light bulbs to protect the filament. \nGreenhouse Effect: Vital to maintaining Earth's climate and life-sustaining warmth. The Sun's heat is trapped within Earth's atmosphere by the greenhouse gases, mostly carbon dioxide, which enters the atmosphere through the planet's natural carbon cycling and increasingly from human-caused pollution. The increase of carbon dioxide gases in the atmosphere is believed to have caused the Earth's average temperature to rise from 60°F to 61°F in the last few decades.\nEarth's Chemical Composition: \n~ Oxygen: 46.6 %\n\n~ Silicon: 27.7%\n\n~ Aluminum: 8.1%\n\n~ Iron: 5%\n\n~ Calcium: 3.6%\n\n~ Sodium: 2.8%\n\n~ Potassium: 2.6%\n\n~ Magne
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You'd have to get everyone on one side of the planet for the weight to do anything. And even then, you'd need a lot more than we have.
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Mass is constant, it does not change based on the gravitational acceleration around it.\n6 kilograms on the Earth would still be 6 kg on the moon.\n\n6 kg on the Earth would have a weight of about 60 Newtons (a little less), but on the moon, it would weight only about 10 Newtons (due to the 1/6 gravity).
83656
You just gave the answer. If he weighs 200 pounds on earth he would weigh 200 pounds on this planet.
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The equatorial radius of earth is 6400 km.\n\nThe angular speed of any object of earth is (2pi/ 24hour).\n\nThe linear speed is (2pi/ 24hour)* R = (2pi/ 24hour)* 6400x 10^3= 465.42m/s.\n\n\na) The speed of 97kg (any object) is 465.42m/s.\n\nb) Since the man is moving along a circle of radius 6400x10^3km, \nthe centripetal force acting on him (toward the center of earth) is m v v / r\n\n= 97 x 465.42 x 465.42 / 6400x 10^3 = 3.28 N \n\n\nc) The acceleration due to gravity at 6400 km is 9.74 m/s^2\n\nThe actual weight of the person is 97 x 9.74 = 944.78 N.\n\nd) If earth were not rotating, the man will press the earth’s surface with a force of 944.78 N.\n\n But, since the earth is rotating, he will press the earth with a force of\n941.5 newton.    (944.78 - 3.28). \n\nThis is the normal reaction acting on the man.
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Of cause, help.\nthe force of gravity is mM/R^2 = 550 N\nThis is for R = 3400km\n\nIn case we take the object up for 12 km, the distance from the centre of the planet will be 3412 N.\n\nand its weight will be mM/(R+12km)^2 = Wn\nWe can find  Wn   just from these two equations:\nfrom the first equation we shall express \nmM = 550(N)xR^2\nthan  Wn = 550 (N) x R^2/(R+12)^2 = 550(N)x3400^2/3412^2 = \n550 (N) x 0.996483^2 = 550(N)x0.9929783 = 546.138 N\nThat is the answer
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Weight is a force, but when something is weighed you actually measure it's mass assuming you are in an environment in which the force of gravity is constant (i.e., you're at a fixed location <radius from the center> on a planet or massive planetary mass).\n\nof course, one's mass can be approximated by W=mg at fixed radii from such planetary masses (like the earth, a massive asteroid, a superdense star, whatever...) if the force of gravity, g, is known at that radius.\n\nThere are 2 possible variables here - mass and the force of gravity.  Which one do you know?  usually an approximation for g is given.  Which one do you measure, then...that would be mass.
72845
use nasa's calculator:\nhttp://www.exploratorium.edu/ronh/weight/
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i'm pretty sure that it's either 3 times greater or 9 times greater coz the gravity on mars is more than that of earth.\nbut i cant decide which one of them is the right one.
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The gravitational pull on the moon is 1/6 that on Earth.  Note that on the Moon only your Weight (Pulling force of planet,etc) changes and not your Mass (Measure of amount of matter in an object).  So if your Mass is 100lb (approx 45kg) on Earth, it would still be 100lb on the Moon.  \nHowever your Weight changes from being 450N (450 newtons using the equation Weight = Mass x gravitational field strength on earth which is taken as 10N/kg) on Earth to 75.2N on the moon (1/6 that on Earth where g = 1.67N/kg)
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The effect which causes weight is actually an acceleration, and in the case of on the surface of Earth it is due to gravity.  Weight may also be caused by acceleration due to motion in a straight line, such as with astronauts during liftoff having their weight "increased".\n\nSo to answer your question depends upon if any acceleration or deceleration is happening during the journey.  If the astronauts are moving toward another planet at a constant speed, then they are completely weightless.  (Well, they will of course be pulled a very tiny amount toward any gravitational sources, but unless they are massive and nearby, the effect will be practically insignificant.)  If they are moving under acceleration, then they will feel "weight" based upon the amount of acceleration.
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to use the gravitational pulls of the earth and the moon
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You are making it needlessly hard on yourself.  The planet has the same mass as the earth, and someone on it weight the same as on earth, therefore the surface of the planet is the same distance from its center as the earth radius, i.e. this planet has the same dimension as earth.\nSo, the radius of the planet is 6,372.797 km, same as earth's.\nYou don't need to compute anything.  The orbital velocity of the planet has nothing to do with it.\nIf the weight was different, you only need to scale it remembering that the weight would vary as an inverse of the square of the radius.
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The astronauts would be in zero-G for almost the entire trip causing the negative effects of weighlessness on their bodies.\n\nThe accereration of the spacecraft would indeed negate the problems with Zero-G by temporarily creating artifical gravity.  The problem lies in the fact that with the limited amount of fuel on the spacecraft, acceleration would only occur while the the engine was on. (a very short time with a chemical rocket)   \n\nOne solution would be to use a a drive that will burn during the entire trip.  One such drive is an ion drive (used on the recent Deep-Space One spacecraft to visit a local asteriod).  This drive would acclerate for most of the trip there creating a small amount of artifical gravity.
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you cant really measure it.  the density and weight of different layers that we cannot reach from the surface makes it very difficult to weigh.