Originally Posted by

**ssadi** The moon can be modeled as a sphere of radius 1740 km and mass 7.44*10^22 kg. A rocket of mass m is fired vertically from the surface of the moon with speed 1.5 km/s. It may be assumed that all forces on the rocket other than the gravitational attraction of the moon can be ignored.

(c) Find the speed of the rocket when t is at a height of 10 km above the surface of the moon.

(d) Find the maximum distance from the moon’s surface reached by the rocket.

My workings:

(M=moon’s mass, m=rocket’s mass, all physical quantities in SI units)

(a)

$\displaystyle F= \frac{GmM}{r^2}$

If r is taken as distance from centre of moon, and denoted by x

$\displaystyle F= {\color{red} -} \frac{GmM}{x^2}$

$\displaystyle \frac{1}{2}mv^2=\int F dx$

$\displaystyle = {\color{red} -} GmM\int\frac{1}{x^2} dx$

So $\displaystyle v^2= \frac{2GM}{x}+k$ Mr F says: I have deleted the negative.

At t=0, v=1.5*10^3, x=1740*10^3

$\displaystyle \therefore k=7877333\frac{1}{3}$ Mr F says: I went to the trouble of finding that, after making the above corrections, k = -3454000.

The rest of the corrections are left for you to make ....

$\displaystyle v^2= -\frac{2GM}{x}+7877333\frac{1}{3} $

$\displaystyle \therefore v=1.51 at x=1750*10^3$

Book’s answer: 1.49

(b)At v=0 x=1243004

But the radius of moon is 1740*10^3

So the answer is absurd, the maximum distance appears negative.

Is there some basic mistake here? Is my method correct? The first answer is quite close but the second one becomes ridiculous:S :help: