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.

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.