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?

______ m

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!

______ m/s

3)

Which of the following would change the minimum velocity needed to make it to the moon? (choose all that apply)

---the mass of the earth

---the radius of the earth

---the mass of the spaceship

Explanation / Answer

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)

2) 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.

3) If you see the ecuation doesn¨t depend of the mass of the spaceship so it just depends on the radius of the earth and the mass of the earth.

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.