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!

3.Which of the following would change the minimum velocity needed to make it to the moon?

Explanation / Answer

1)
potential energy at the surface of the earth:
U = - GmM/R with R = earth radius, M = earth mass, m =
potential energy at distance r from the earth center
U = -GmM/r
potential energy difference = -GmM(1/r - 1/R)
initial kinetic energy = potential energy difference
1/2 mv^2 = -GmM(1/r - 1/R)
1/2*v^2 = -GM(1/r - 1/R)
1/2*5534^2 = - 6.67*10^-11*5.97*10^24(1/r - 1/(6.38*10^6))
r = 8.45*10^6 m
(= distance from earth center that can be reached. Do not know if the potential energy gain due to moon mass should be included or not)
---------
potential energy at moon surface due to earth mass:
U = -GmM(e)/(3.84*10^8 - 1.74*10^6) with M(e) = earth mass
net potential energy at moon surface due to earth = (- GmM(e)(1/(3.827*10^8)- 1/(6.38*10^6)) (must be subtracted from initial kinetic energy)
potential energy at moon surface due to moon mass
U = -GmM(m)/(1.74*10^6) with M(m) = moon mass
potential energy due to moon at earth surface:
U = -GmM(m)/(3.827*10^8 - 6.38*10^6)
net potential energy due to moon at moon surface
= - GM(m)m(1/(3.763*10^8) - 1/(1.737*10^6)) (this adds to the final kinetic)

net kinetic energy at the surface of the moon due to masses of earth and moon and initial kinetic energy on earth surface:
Ek(final) is
1/2* mv^2 = 1/2mv0^2 - (- GmM(e)(1/(3.827*10^8)- 1/(6.38*10^6)) +
(- GM(m)m(1/(3.763*10^8) - 1/(1.737*10^6)))
1/2* v^2 = 1/2 v0^2 - (-GM(e)(1/(3.827*10^8)- 1/(6.38*10^6)))
+ (-GM(m)(1/(3.763*10^8) - 1/(1.737*10^6)))
0.5v^2 = 0.5*11068^2 - (-6.67*10^-11*5.97*10^24(1/(3.827*10^8)- 1/(6.38*10^6)))
+(-6.67*10^-11*7.36*10^22(1/(3.763*10^... - 1/(1.737*10^6)))
---> v = 2319.63 m/s (= final velocity at moon surface)

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

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