Question

An unmanned spacecraft is in a circular orbit around the moon, observing the lunar surface from an altitude of 55.0km . To the dismay of scientists on earth, an electrical fault causes an on-board thruster to fire, decreasing the speed of the spacecraft by 15 m/s. If nothing is done to correct its orbit, with what speed (in km/h) will the spacecraft crash into the lunar surface? .

Answer 1

You need the mass of the moon and the radius of the moon.

Then work out V^2 = G . M / R

Get V . Take away 15 m/s .

Calculate new KE of the satellite ( it will have an "m" in it )

Calculate the height energy that the satellite has ( m .g . h ) - the h is 55 km and the "g" is the surface acceleration due to gravity on the Moon. You can get away with just taking it as the surface gravity because 55 km is not far in comparison to the radius of the Moon.

Add this energy to the KE that the satellite has - this is the total energy the satellite has on impact. ( it still has an "m" in it ).
This = 1/2 . m . v^2
where v is the speed of impact. ( and the "m's" cancel )

Answer 2

I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else)

An unmanned spacecraft is in a circular orbit around the moon,observing the lunar surface from an altitude of50.0 (see Appendix F). To the dismay of scientists on earth,an electrical fault causes an on-board thruster to fire, decreasingthe speed of the spacecraft by 20.0
m/s
. If nothing is done to correct its orbit, with whatspeed (in km/h) will the spacecraft crash into the lunarsurface?

answer

12.63: and basic concepts of orbits. E = K + U, or-GM-1V2-GM, where E,K and U are the energies per unit mass, v is the circular orbital velocity of 1655 m/s at the lunicentric distance of 1.79 x10 m. The total energy at this distance is -1.37x106 J/Kg. When the velocity of the spacecraft is reduced by 20 m/s, the total energy becomes One can solve this problem using energy conservation, units of J/kg for energy, (6.673 x10-11 N m2 /kg2) (7.35 x1022 kg) (1.79x10 m) E(1655 m/s 20 m/s)2 - 2 2 or 1.40x106 J/kg. Since E we can solve for a,a 1.748x10 m, the semi major axis of the new elliptical orbit. The old distance of 1.79x106 m is now the apolune distance, and the perilune can be found from a1.706x10 m. Obviously this is less than the radius of the moon, so the spacecraft crashes ! At the surface, U-_ or. U--281 8 × 106 J/kg Since the total energy at the surface is -1.40x106 J/kg, the kinetic energy at the surface is 1.415x 106 J/kg. So,y-1.415x 106 J/kg, or v= 1.682x 103 m/s= 6057 km/h.

Answer 3

h = height of spacecraft =55 km
Rm = radius of the moon = 1738 km
r = radial distance to satellite = Rm + h = 1738 + 55 = 1793 km = 1.793x10^6 m
g = acceleration of gravity on the moon = 1.6 m/s^2
M = mass of the moon = 7.36 x10^22 kg
G = gravitational constant = 6.673 x 10-11 N

Answer 4

h = height of spacecraft = 55 km
Rm = radius of the moon = 1738 km
r = radial distance to satellite = Rm + h = 1738 + 55 = 1793 km = 1.793x10^6 m
g = acceleration of gravity on the moon = 1.6 m/s^2
M = mass of the moon = 7.36 x10^22 kg
G = gravitational constant = 6.673 x 10-11 N