Oh well, as a first guess since you did not say I will assume g moon = g earth/6
= 9.8/6 = 1.63 m/s^2
so
a = -1.6
v = Vi - 1.6 t
(2/5)(1210) = 1210 - 1.6 t
1.6 t = 1210 (3/5)
t = 454 seconds
h = Vi t - ((1/2)(1.6) t^2
= 384,447 meters
A projectile is launched vertically from the surface of the Moon with an initial speed of 1210 m/s. At what altitude is the projectile's speed two-fifths its initial value?
8 answers
the mass of the moon is M=7.35e22
and the radius= 1.738 e6
I used the formula
1/2m(2/5vi)^2 - GMm/h+R= 1/2vi^2-GMm/R
Where little m cancels out and solved for h. But my answer and the answer you just gave me was marked wrong.
So you have any other suggestions?
and the radius= 1.738 e6
I used the formula
1/2m(2/5vi)^2 - GMm/h+R= 1/2vi^2-GMm/R
Where little m cancels out and solved for h. But my answer and the answer you just gave me was marked wrong.
So you have any other suggestions?
well, what did you calculate for g moon? If it is not something like 9.81/6 it is wrong, but the 1/6 of earth gravity is a rough approximation.
6.67e-11
Your way I get
(1/2)(21/25) vi^2 = G M [1/R - 1/(R+h) ]
21/50 (1210)^2 = 6.67*10^-11 * 7.35*10^22
[ 1/1.74*10^6 - 1/(1.74*10^6+h) ]
.615*10^6=49*10^11 (.575*10-6 - z)
where
z = 1/(1.74*10^6+h)
.1255*10^-6 = .575*10^-6 - z
z = .449*10^-6
1/z = 2.22*10^6 = 1.74 *10^6 + h
h = .485 *10^6
485,000 meters
check our arithmetics
(1/2)(21/25) vi^2 = G M [1/R - 1/(R+h) ]
21/50 (1210)^2 = 6.67*10^-11 * 7.35*10^22
[ 1/1.74*10^6 - 1/(1.74*10^6+h) ]
.615*10^6=49*10^11 (.575*10-6 - z)
where
z = 1/(1.74*10^6+h)
.1255*10^-6 = .575*10^-6 - z
z = .449*10^-6
1/z = 2.22*10^6 = 1.74 *10^6 + h
h = .485 *10^6
485,000 meters
check our arithmetics
using your figures
g moon = F/m = G M/r^2
= 6.67^10^-11*7.35*10^22 /3.03*10*12
=16.17*10^-1 = 1.67 m/s^2
I guessed 9.8/6 = 1.63 so I would be a little off
g moon = F/m = G M/r^2
= 6.67^10^-11*7.35*10^22 /3.03*10*12
=16.17*10^-1 = 1.67 m/s^2
I guessed 9.8/6 = 1.63 so I would be a little off
okay, I got it thanks !
Great !