A meteor is falling towards the Earth. If Mars and radius of the Earth are 6×10^24 kg and 6.4×10^3 km respectively. Find the height of meteor from the Earth's surface where its acceleration due to gravity becomes 4m/s^2. please explain it and solve it
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Answer:
It is a slightly tricky question. By the way the radius of the earth is 6.371 • 10^6 meters ! let’s use this figure as it is more correct to three decimal places. What you really want to find out is at what distance from the center of the earth which includes the radius of the earth plus the height above the earth’s surface that the acceleration due to gravity will be 4m/s^2. As you know on and near the surface of the earth the acceleration due to gravity is 9.81m/s^2. So it has to be very high up above the earth’s surface, way above the atmosphere which ascends up to 300 miles(480,000 meters) above the earth. The equation for the acceleration due to gravity derived from Newton’s equation of universal gravitation is shown as follows :- g = G•M/D^2, and here D = d + h where d is the radius of the earth, and h is the height above the earth where the meteor in the question is. So the above equation can be rewritten as : g = G•M/(d+h)^2. It is the h, the height above the earth’s surface that we want to calculate here. This ‘gravity-mediated’ acceleration of free fall is independent of the mass of the falling object as mass of the object cancels out from both sides of the equation giving you above. So solving the equation mathematically gives us what is shown below : (d+h) = (G•M/g)^1/2 which gives us : h = {(G•M/g)^1/2} - d which is the final mathematical expression for the height at which the meteor is above the earth’s surface. Remember the g value at that height h is given as 4m/s^2(not 9.81 m/s^2 which is on the surface of the earth). Plugging in the values will give us h = (10.005 • 10^6 m) - (6.371 • 10^6 m) which is a height of 3.634 • 10^6 m or 3,634,000 meters above the earth’s surface. Each mile is 1609.344 meters, so that is 2,258.1 miles above the earth’s surface which is way up in space ! The atmosphere ascends up to 300 miles(480 km) above the earth, so this meteor is up there in space way above the earth’s atmosphere, at least 1,942 miles above the upper limit of the earth’s atmosphere. As it descends more and more the acceleration due to gravity will go up more and more with the value of D(d+h)shrinking, as h will dwindle more and more until it starts to enter earth’s atmosphere when h becomes 300 miles or so. Upon entry into the atmosphere it will start to get heated up greatly due to friction, and a lot or all of it will burn up in the atmosphere during its descent before it can make landfall…