How Far Will the Rocket Be from Earth?Date: 06/22/2001 at 11:47:58 From: Elaine Chesebro Subject: Algebra Hello, Can you explain the process you would use in solving this problem? A rocket leaves the earth for the sun at a speed of 28,800 mph. At the same time, a photon of light leaves the sun for the earth. If we assume they are on flight paths that allow them to meet, how far will the rocket be from earth when when the photon of light hits it? (Assume a distance of 93,000,000 miles to the sun and round your answer to the nearest thousand miles.) (A rocket launched from the earth will gradually build up speed. For this problem, however, we will assume that the rocket travels at 28,800 mph the instant it leaves the earth.) Can you help? Thanks! Date: 06/22/2001 at 12:33:44 From: Doctor Ian Subject: Re: Algebra Hi Elaine, Whoever made up this problem also seems to be assuming that the earth and sun are standing still, and that the way you send a rocket to the sun is to launch it straight at the apparent location of the sun, rather than in an elliptical orbit (which is actually how you would have to do it). So let's forget about the incomplete 'context' of the problem, and concentrate on the math. The distance traveled by the photon and the distance traveled by the rocket have to add up to 93 million miles: rate(rocket) * time(rocket) + rate(photon) * time(photon) = 93 million miles Since the photon and the rocket leave at the same time, time(rocket) = time(photon) = t so rate(rocket) * t + rate(photon) * t = 93 million miles t * (rate(rocket) + rate(photon)) = 93 million miles You're given the rate of the rocket. You can look up the rate of the photon. Make sure you use values that have the same units. So you can solve for t. Once you know t, you can find out how far the rocket moves: distance(rocket) = rate(rocket) * t That's how I would solve it... unless I was in a hurry. :^D In that case, I would do this. Let r be the rate of the rocket, and kr be the rate of the photon - that is, assume that the speed of the photon is k times greater than the speed of the rocket. Let t be the common time of travel, and d be the distance from the earth to the sun. Then rt + krt = d rt(1 + k) = d rt = d/(1 + k) So the distance traveled by the rocket would be 93 million miles ---------------------------- 186,000 miles/second 1 + -------------------- 28,800 miles/hour which is really the same thing as 28,800 miles/hour 93 million miles * ------------------------------------- 28,800 miles/hour + 186,000 miles/sec The basic idea is that if the rocket is contributing some fraction of the combined speed, it will contribute the same fraction of the combined distance. Does that make sense? Note that you still have to convert both speeds to either miles/second or miles/hour. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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