# Math Help - Finding the initial velocity

1. ## Finding the initial velocity

Estimate the initial velocity in meters/s that needs to be given to a spacecraft moving straight away from earth to make it reach the lunar orbit at 383000 km away from the centre of the earth.
The initial height of the spacecraft is H km above the surface of the earth. The radius of the earth is 6370 km.

H[km] = 6220;

2. ## Work done by gravitational force

Hello m_i_k_o
Originally Posted by m_i_k_o
Estimate the initial velocity in meters/s that needs to be given to a spacecraft moving straight away from earth to make it reach the lunar orbit at 383000 km away from the centre of the earth.
The initial height of the spacecraft is H km above the surface of the earth. The radius of the earth is 6370 km.

H[km] = 6220;

Here are the general principles to use:

• The gravitational pull exerted by the earth on the spacecraft is inversely proportional to the square of its distance from the centre of the earth. So if this force is $F$ N, the mass of the spacecraft is $m$ kg, and its distance is $x$ m from the centre of the earth,

$F = \frac{km}{x^2}$

• At the earth's surface this pull is equal to the weight of the spacecraft; i.e. when $x = 6370\times10^3, F = 9.8m$ N. This enables you to find $k$.

• When the spacecraft moves outwards from the earth from $x=x_0$ to $x=x_1$, the work done on it by the earth's gravitational pull is:

$\int_{x_0}^{x_1}-Fdx$

• Then use the Work-Energy Principle to find the initial velocity knowing that the final velocity is zero.

If it still doesn't work out, show us your working so far, and we'll have another look at it.

Granddad, could you please elaborate on dot points 3 and 4? I get the integral of F to be (-2km)/x^3 but what am I to do with the mass??? please help

4. Hello StarkMate

Welcome to Math Help Forum!
Originally Posted by StarkMate
Granddad, could you please elaborate on dot points 3 and 4? I get the integral of F to be (-2km)/x^3 but what am I to do with the mass??? please help
I think you've differentiated.
$\int -\frac{km}{x^2}\; dx = -\frac{kmx^{-1}}{(-1)}$
$=\frac{km}{x}$
Now you need to evaluate this integral between the limits $x_0$ and $x_1$ (where $x_0 = (6370+6220)\times 10^3$ and $x_1 = 383000\times 10^3$).

This, when you've worked out the value of $k$, gives an expression in terms of $m$, which is equal to the loss of KE of the spacecraft. This will be equal to $\tfrac12mv^2$, where $v$ is the velocity required. The $m$'s then cancel, leaving an expression for $v$.