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It's Only a Minus Sign

Solve these differential equations to see how a minus sign can change the answer

Out in Space

Age 16 to 18
Challenge Level
Thank you to Tarang James of Kerang Technical High School and Andre from Tudor Vianu National College, Romania for your solutions to this problem.

First I shall prove the formula: $$a = v{dv\over dx}\quad (1)$$ I shall write the formula for acceleration ${dv\over dt}$ as $$a = {dv\over dt} = {dv\over dx}.{dx\over dt} = v. {dv\over dx}$$

From the problem, I know that $a = -c/x^2$, as vectors $a$ and $x$ have opposite signs. Using this and relation (1), I obtain: In this example $${dv\over dt} = v.{dv\over dx} = -{c\over x^2}$$ and hence $$\int v dv = \int {-c\over x^2} dx\;,$$ so $${v^2\over 2}= {c\over x} + k$$ and we are given $c=4\times 10^5$, and $v=10$ when $x=10^4$ so $$ k = -40 + 50 = 10.$$ Hence when $v=5$ we have $$12.5 = {c\over x} + 10$$ which gives $x = {4\times 10^5 \over 2.5}= 160,000$. So the space craft moves at $5 \; \text{ km}$ per second when it is $160,000 \; \text{km}$ from the Earth.

I solved the problem in two ways, as above using the relation just proved, and using conservation of energy. The methods are essentially the same and the constant given as $c$ combines the gravitational constant $G$ and the mass of the Earth.