A moon lander is orbiting the lOOo emm . meron at an altitud 04 ecyease \'ts spead so as to duse eriod lati hat 2erce must it dectease its 3\"ay.tw mron\'t Çatau. ?n-half
Solution
Let A = the altitude above the moon = 1000 km = 1 x 10^6 m
Let R = the radius of the moon = 1.74 x 10^6 m
Let G = the gravitational constant = 6.67 x 10^-11
Let M = the mass of the moon = 7.35 x 10^22 kg
Let V = the tangential velocity of the lander
The above is the list of data that you need; it can be found using any search engine.
The centripetal acceleration equation is:
a = V²/(R + A)
The equation for acceleration due to lunar gravity (g) is:
g = G(M)/(R + A)²
In a stable obit, the acceleration due to gravity is equal to the centripetal acceleration:
V²/(R + A) = G(M)/(R + A)²
V² = G(M)/(R + A)
V = ?{G(M)/(R + A)} = 1337.62 m/s
To find the new velocity (V1) so that the orbit decays to just graze the surface, let A?0
V1 = ?{G(M)/(R)} = 1678.54 m/s
There is, however, no way to assure that the transition will take one-half period to occur without doing multiple velocity changes.
The percentage change is:
% = 100(V1 - V)/V
= 25.5%
.
A moon lander is orbiting the lOOo emm - meron at an altitud 04 ecyeas.docx
1. A moon lander is orbiting the lOOo emm . meron at an altitud 04 ecyease 'ts spead so as to duse
eriod lati hat 2erce must it dectease its 3"ay.tw mron't Çatau. ?n-half
Solution
Let A = the altitude above the moon = 1000 km = 1 x 10^6 m
Let R = the radius of the moon = 1.74 x 10^6 m
Let G = the gravitational constant = 6.67 x 10^-11
Let M = the mass of the moon = 7.35 x 10^22 kg
Let V = the tangential velocity of the lander
The above is the list of data that you need; it can be found using any search engine.
The centripetal acceleration equation is:
a = V²/(R + A)
The equation for acceleration due to lunar gravity (g) is:
g = G(M)/(R + A)²
In a stable obit, the acceleration due to gravity is equal to the centripetal acceleration:
V²/(R + A) = G(M)/(R + A)²
V² = G(M)/(R + A)
V = ?{G(M)/(R + A)} = 1337.62 m/s
To find the new velocity (V1) so that the orbit decays to just graze the surface, let A?0
V1 = ?{G(M)/(R)} = 1678.54 m/s
2. There is, however, no way to assure that the transition will take one-half period to occur without
doing multiple velocity changes.
The percentage change is:
% = 100(V1 - V)/V
= 25.5%
