# Out in Space

A space craft is ten thousand kilometres from the centre of the
Earth moving away at 10 km per second. At what distance will it
have half that speed?

## Problem

Show that the acceleration ${dv\over dt}$ of a particle moving in a straight line can be written, in terms of its velocity $v$ and its displacement $x$ from a point of the line, in the form $v{dv\over dx}$.

At a distance $x\;\text{km}$ from the centre of the Earth the gravitational acceleration in $\text{km s}^{-2}$ is given by the formula $\frac{c}{x^2}$ where $c=4 \times 10^5$. If a space craft $10^4 \; \text{km}$ from the centre of the Earth is moving directly away from it at a speed of $10 \; \text{km s}^{-1}$ at what distance will it be moving with half that speed?

## Getting Started

Use the first part of the question to do the second part.

If you can separate the variables you can write the expression in the form $$\int f(x) dx = \int f(v) dv$$ and integrate to find the solution.

Take care over signs, acceleration is towards the Earth and displacement and velocity are measured in the opposite direction.

If you can separate the variables you can write the expression in the form $$\int f(x) dx = \int f(v) dv$$ and integrate to find the solution.

Take care over signs, acceleration is towards the Earth and displacement and velocity are measured in the opposite direction.

## Student Solutions

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.

## Teachers' Resources

Students: Don't be deterred if you have not learnt the method of solving differential equations by separation of variables because all you have to do is to rearrange the equation in the form $$\int f(v) dv = \int f(x) dx$$ so that you can integrate both sides and the rest is easy integration.

Teachers: This is an example you might use to launch into work on learning the method of separation of variables.