Associated Topics || Dr. Math Home || Search Dr. Math

### 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/
```
Associated Topics:
High School Basic Algebra

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search