What Is the Gravitational Potential on the Moon's Surface?

AI Thread Summary
The discussion revolves around calculating the gravitational potential on the Moon's surface using the formula V = -G[Mm/Rm + Me/(D - Rm)]. The user initially attempted to use V = -GMm/Rm but was confused about why the Earth's mass is added rather than subtracted. It was clarified that the Earth's gravitational potential also contributes positively to the total potential experienced on the Moon's surface. The assumption is made that the calculation pertains to the Moon's light side, where the Earth's influence is more significant. Understanding these contributions is crucial for accurately determining the gravitational potential in such scenarios.
crossfacer
Messages
20
Reaction score
0

Homework Statement



Given:
G = 6.67*10^ -11 Nm^2kg^-2
Radius of Moon = Rm = 1.73*10^6 m
Radius of Earth = Re = 6.37*10^6 m
acceleration due to gravity = g = 9.8m/s
Mass of moon = Mm
Mass of Earth = Me
Distance between the centre of the Earth and the centre of the moon = D = 3.8*10^8 m

Find the gravitational potentials on the moon's surface.

The Attempt at a Solution



V = - GMm/Rm
V = - g * Re^2 / Rm
V = - 9.8 * (6.37*10^6)^2 / 1.73*10^6
V = - 2.30*10^8 J/kg

The solution said that V should be
-G* [ Mm/Rm + Me/(D - Rm) ]
I don't understand why it should not be "Mm/Rm - Me/(D - Rm)" :confused:

Thank you very much!
 
Physics news on Phys.org
Because the Earth contributes a potential too. I guess you are supposed to assume that the problem wants the potential on the light side of the moon.
 
Thank you! In the past I thought that some potential of Earth will be canceled out by potential of the moon :-p
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Modeling a kongming lantern (sky lantern)
Replies
6
Views
2K
Potential Energy and raising a satellite from Earth into a Circular Orbit
Replies
5
Views
2K
Variation of escape velocity with equal surface gravity
Replies
7
Views
1K
Work and Potential Energy: To the Moon physics problem
Replies
4
Views
5K
Gravitational Field Strength at the equator and poles
Replies
23
Views
3K
Gravitational Potential Energy Project Thor
Replies
8
Views
2K
Where between the Moon and the Earth is the gravitational potential=0 ?
Replies
21
Views
3K
Calculating Tidal Range (Gravitation)
Replies
12
Views
2K
Finding distance between Earth and Moon in gravitational fields
Replies
5
Views
2K
Difference in Gravitational Potential on the Moon
Replies
15
Views
2K

Hot Threads

Recent Insights

Back
Top