# Trip to the moon

1. Sep 25, 2010

### hime

Solved:Trip to the moon

1. The problem statement, all variables and given/known data

You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the earth with enough speed to make it to the moon. Some information that will help during this problem:
mearth = 5.9742 x 10^24 kg
rearth = 6.3781 x 10^6 m
mmoon = 7.36 x 10^22 kg
rmoon = 1.7374 x 10^6 m
dearth to moon = 3.844 x 10^8 m (center to center)
G = 6.67428 x 10^-11 N-m2/kg2
The minimum speed needed to reach the moon is vmin = 11068 m/s. If you leave the surface of the earth at this speed, how fast will you be moving at the surface of the moon?

2. Relevant equations

Fgrav=G*m1*m2/r^2
PE earth=PE moon + KE moon ??
PE total = -GMe*m/ D-Re - GMm*m/D-Rm where D is the distance between centers of earth and moon, m is the mass of the person, Mm is the mass of the moon, and Me is the mass of the earth

3. The attempt at a solution
The velocity cannot be zero at the surface of the moon. Although there is a point along the way that your speed approaches zero, as the spaceship falls to the moon it speeds up.

Last edited: Sep 26, 2010
2. Sep 25, 2010

### nickjer

This sounds like a conservation of energy problem and you are on the right track, but what happened to KE_earth? Remember, they said he leaves Earth at a velocity v_min = 11068 m/s.

3. Sep 25, 2010

### hime

you mean KE earth+PE earth=PE moon + KE moon ??

I plugged in numbers but still got 11068 m/s when i solved for Vmoon which is the vmin they gave us in the problem...that just does not make sense.

If I want to use the kinematics equation: V^2=Vi^2=2a(Xf-Xi), I don't know the height that the object will be above the moon before it falls down to the surface.

Is there a way to do this problem using center of mass concept?

4. Sep 25, 2010

### nickjer

You cannot use any of your kinematics equations since those only apply to constant acceleration. But the gravitational acceleration changes as the distance changes.

5. Sep 25, 2010

### hime

KE earth+PE earth=PE moon + KE moon

.5*m*ve^2 + m*G*he = m*G*hm + 0.5*m*vm^2

i took he and hm as the radius of the earth and moon respectively.

m cancels on both sides so the equation becomes and solving for vm:
.5*ve^2 + G*he = G*hm + 0.5*vm^2
.5*11068^2+6.67428e-11*6.3781e6 = .5*vm^2 + 6.67428e-11*1.7374e6

Vm = 11068 m/s

6. Sep 25, 2010

### nickjer

g is not the same on both celestial bodies. You also should not be using PE = mgh. That is an approximation used for the surface of the earth where 'g' is 9.8 m/s^2. You should be using:

$$PE = -\frac{GMm}{r}$$

where M = mass of planet, m = mass of object, and r = radial distance of object from center of planet.

7. Sep 25, 2010

### hime

ok..i substituted the PE that you stated and got the following equation:

KE earth+PE earth=PE moon + KE moon

.5*m*ve^2 +G*Me*m/re = 0.5*m*vm^2+G*Mm*m/rm

Plugging all values I got vm=1.11818e9 m/s which is wrong answer why?

8. Sep 26, 2010

### hime

I found the old thread from physics forums and it discusses the same problem as me

https://www.physicsforums.com/archive/index.php/t-206632.html

Let's say there is a point P between Earth and the Moon where the velocity will reach 0m/s.
The escape velocity of the earth is just the velocity needed for the object to reach point P.
From Point P, the force of the moon's gravity will take over the object and pull it towards its surface. To calculate the Vfinal, we can use conservation of energy from Point P to moon's surface.

The vfinal = 2.28km/sec using conservation of energy.

9. Sep 26, 2010

### hime

Thanks nickjer for helping me..:)

10. Sep 27, 2010

### jdgallagher14

do you go to u of i? physics 211? this is my homework problem. lol.