# Math Help - Stuck on a mechanics sum in moon's gravity:help:

1. ## Stuck on a mechanics sum in moon's gravity:help:

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

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

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

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

So $v^2= -\frac{2GM}{x}+k$

At t=0, v=1.5*10^3, x=1740*10^3
$\therefore k=7877333\frac{1}{3}$

$v^2= -\frac{2GM}{x}+7877333\frac{1}{3}$
$\therefore v=1.51 at x=1750*10^3$
(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:

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

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

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

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

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

At t=0, v=1.5*10^3, x=1740*10^3
$\therefore k=7877333\frac{1}{3}$ 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 ....

$v^2= -\frac{2GM}{x}+7877333\frac{1}{3}$
$\therefore v=1.51 at x=1750*10^3$
(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:
Since gravity is the only force acting on the rocket, the fact that you get a speed at 10km above the surface that is greater than the speed at the surface should have told you that there's an error.

It's not good to say that 1.51 km/sec is close enough to 1.49 km/sec. It's very wrong because it's greater than 1.5 km/sec (the speed at the surface). You should expect that the rocket slows down, that is, loses kinetic energy as its distance from the moon increases, that is, gains gravitational potential energy.

See the red for some of the corrections - the ripple effect is left for you to deal with. I get the books answer for the first question.

3. Originally Posted by mr fantastic
Since gravity is the only force acting on the rocket, the fact that you get a speed at 10km above the surface that is greater than the speed at the surface should have told you that there's an error.

It's not good to say that 1.51 km/sec is close enough to 1.49 km/sec. It's very wrong because it's greater than 1.5 km/sec (the speed at the surface). You should expect that the rocket slows down, that is, loses kinetic energy as its distance from the moon increases, that is, gains gravitational potential energy.

See the red for some of the corrections - the ripple effect is left for you to deal with. I get the books answer for the first question.
I should have recognized the paradox
Thanks for the negative uprooting effort.