You plan to take a trip to the moon. Since you do not have a traditional spacesh

Question

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 1024 kg

rearth = 6.3781 x 106 m

mmoon = 7.36 x 1022 kg

rmoon = 1.7374 x 106 m

dearth to moon = 3.844 x 108 m (center to center)

G = 6.67428 x 10-11 N-m2/kg2

1) On your first attempt you leave the surface of the earth at v = 5534 m/s. How far from the center of the earth will you get?

2) Since that is not far enough, you consult a friend who calculates (correctly) the minimum speed needed as 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? Hint carefully write out an expression for the potential and kinetic energy of the ship on the surface of earth, and on the surface of moon. Be sure to include the gravitational potential energy of the earth even when the ship is at the surface of the moon!

(Please inlude step-by-step solution or even an online video link that can teach me this as well)

Explanation / Answer

SOLUTION=

The gravitational force is:

F = G M m / r²

And it sits in a potential field given by U(r), where F = - U is the gradient of this potential:

U = - G M m / r

You start off with kinetic energy given by K = ½ m v², and it disappears into this potential energy reservoir, so that the change in kinetic energy (-K, because it starts at K and goes to 0) is the negative of the change in potential energy (proportional to (1/r) (1/r)):

-½ m v² = G M m * [(1/r) (1/r)]

As you can see, your mass is unimportant and divides out, so that you instead just get:

1/r = 1/r v² / (2 G M)
r = 1 / [(1/r) v² / (2 G M)]

Using r as the radius of the earth, v as 5534 m/s, and G and M as given, I get that r = 8448 km, in agreement with your number if your number is in meters. (Please do not forget units; they are one of the most important things that you need to know in physics. Any physical quantity comes with three things: a number, a unit, and an uncertainty. If you don't know all of these, then you don't know anything about that physical quantity.)

Your friend's calculations roughly agree with calculating v for r = , the so-called escape velocity of Earth:

0 = 1/r v² / (2 G M)
v = ( 2 G M / r ) = 11.18 km / s.

It's maybe a little less because you don't have to go all the way out to infinity (just out to the Moon), and the Moon's own gravity helps you a little. A proper derivation would look for the place where the potential is highest, which is where the forces are equal:

G M / r² = G m / (D r)²
r = D (M (M m)) / (M m)

...and would then evaluate the gravitational potential -GM/r - Gm/(D r) at that point, to figure out how much kinetic energy you'd need. But the dominant terms come from the Earth, simply because it is so darn massive and the Moon is so darn far away.

It's another energy balance equation, though: energy to start with is the same as energy that you end with. Suppose that we start a distance r from the Earth and end a distance r from the Moon, then the energy balance gives:

½ v² G M / r G m / (D r) = ½ v² G M / (D r) G m / r

...where m is the moon's mass. (Again, your own mass is unimportant and has been divided out.)

One simple limit takes D and ½ v² G M / r (the escape velocity equation), to yield:

½ v² G M / r
v ( 2 G M / r ) = 2377 m/s.

... but of course you can do the whole solution as you like. (For a physicist, it's enough to know that it's on the ballpark of 2 km/s -- this is already enough to tell you that you will need to bring a lot of fuel to slow down as you approach the Moon's surface.)

1) On your first attempt you leave the surface of the earth at v = 5534 m/s. How far from the center of the earth will you get?
ANSWER = 8448692
2)Since that is not far enough, you consult a friend who calculates (correctly) the minimum speed needed as 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?
m/s
ANSWER = 2280.

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