Difference in Gravitational Potential on the Moon

[AN630078]

Homework Statement

Hello, I have been learning about gravitational fields recently and am awash with formula to calculate the gravitational force, field strength and potential. To be perfectly candid I have not entirely secured my grasp on the topic yet and have been practising some questions to refine my comprehension. The question below concerns finding the gravitational potential on the Moon. I have endeavoured to solve it comprehensively but I am still a little uncertain. Could anyone offer some further guidance to areas where I may perhaps be stumbling a little.

The Moon has a gravitational field near its surface one sixth of that of the Earth. Calculate:
1. The weight of a man of mass 90 kg on the Moon.
2. The gravitational potential energy he would gain in climbing a 50 m hill on the Moon.
3. The difference in gravitational potential between the top and bottom of the hill.

Relevant Equations

W=mg
∆Ep=mg∆h
V grav=-Gm/r

1. Since the gravitaional field strength is 1/6 of that on Earth:
W=mg
W=90*9.81/6
W=90*1.635
W=147.15 ~ 147 N

2. ∆Ep=mg∆h
∆Ep=90*1.635*50
∆Ep=7357.5 J

I do not now whether this method would be suitable and if I should have instead used the formula for gravitaional Potential, V grav=-Gm/r?

3. This is where I am most confused, I contemplated using V grav=-Gm/r but I do not know the mass or the radius of the moon to use this formula?
I am sorry I am just really stuck here.

 

[PeroK]

Homework Statement:: Hello, I have been learning about gravitational fields recently and am awash with formula to calculate the gravitational force, field strength and potential. To be perfectly candid I have not entirely secured my grasp on the topic yet and have been practising some questions to refine my comprehension. The question below concerns finding the gravitational potential on the Moon. I have endeavoured to solve it comprehensively but I am still a little uncertain. Could anyone offer some further guidance to areas where I may perhaps be stumbling a little.

The Moon has a gravitational field near its surface one sixth of that of the Earth. Calculate:
1. The weight of a man of mass 90 kg on the Moon.
2. The gravitational potential energy he would gain in climbing a 50 m hill on the Moon.
3. The difference in gravitational potential between the top and bottom of the hill.
Relevant Equations:: W=mg
∆Ep=mg∆h
V grav=-Gm/r

1. Since the gravitaional field strength is 1/6 of that on Earth:
W=mg
W=90*9.81/6
W=90*1.635
W=147.15 ~ 147 N

2. ∆Ep=mg∆h
∆Ep=90*1.635*50
∆Ep=7357.5 J

I do not now whether this method would be suitable and if I should have instead used the formula for gravitaional Potential, V grav=-Gm/r?

3. This is where I am most confused, I contemplated using V grav=-Gm/r but I do not know the mass or the radius of the moon to use this formula?
I am sorry I am just really stuck here.

Near the surface of a large object like the Earth or the Moon the gravitational force is almost constant. If $h$ is small compared to the radius of the planet, then you can use $mgh$ for gravitational PE. Where $g$ is the surface gravity and $h$ is the height above the surface.

But, if the range of distances is not small, then you must use $-\frac{GMm}{r}$, as this gives the correct variation in the gravitational force, where $r$ is the distance from the centre.

 

[AN630078]

Near the surface of a large object like the Earth or the Moon the gravitational force is almost constant. If $h$ is small compared to the radius of the planet, then you can use $Mgh$ for gravitational PE. Where $g$ is the surface gravity and $h$ is the height above the surface.

But, if the range of distances is not small, then you must use $-\frac{GM}{r}$, as this gives the correct variation in the gravitational force, where $r$ is the distance from the centre.

Thank you for your reply I very much appreciate it. So in the case of question 3, the height above the surface of the Moon being 50m would be considered small when compared to the radius of the planet so I could use Ep=mgh.

How would I progress from here though?

 

[PeroK]

Thank you for your reply I very much appreciate it. So in the case of question 3, the height above the surface of the Moon being 50m would be considered small when compared to the radius of the planet so I could use Ep=mgh.

How would I progress from here though?

Progress in what way?

 

[AN630078]

Progress in what way?

To progress to find the difference in gravitational potential between the top and bottom of the hill. Would it be that at the bottom of the hill, all of the enregy is potential energy, meaning KE=0.
Would I use my result for the potential energy the astronaut has gained from traveling to the top of the hill, i.e. ∆Ep=7357.5 J and substract the Ep at the bottom.
At the bottom;
∆Ep=90*1.635*0m=0 sorry I must be doing something wrong. I realize I am making very simple mistakes I just cannot quite get my head around this problem?

 

[PeroK]

To progress to find the difference in gravitational potential between the top and bottom of the hill. Would it be that at the bottom of the hill, all of the enregy is potential energy, meaning KE=0.
Would I use my result for the potential energy the astronaut has gained from traveling to the top of the hill, i.e. ∆Ep=7357.5 J and substract the Ep at the bottom.
At the bottom;
∆Ep=90*1.635*0m=0 sorry I must be doing something wrong. I realize I am making very simple mistakes I just cannot quite get my head around this problem?

The answers to 2) and 3) must be the same surely? If you gain a certain amount of PE in climbing a hill, then that is the difference in PE between the bottom and the top.

What else could you possibly calculate?

 

[AN630078]

The answers to 2) and 3) must be the same surely? If you gain a certain amount of PE in climbing a hill, then that is the difference in PE between the bottom and the top.

What else could you possibly calculate?

Thank you for your reply. Well that is what I thought, maybe my answer for 2 is wrong? Or maybe since they would essentially be the same thing that is why I thought the question may be hinting at the difference in gravitaional potential of the Moon, but I realize from your statement that the distance climbed is too small really to find such a change.

I am really struggling to answer this suitably!? Do you have nay suggestions?

 

[PeroK]

Thank you for your reply. Well that is what I thought, maybe my answer for 2 is wrong? Or maybe since they would essentially be the same thing that is why I thought the question may be hinting at the difference in gravitaional potential of the Moon, but I realize from your statement that the distance climbed is too small really to find such a change.

I am really struggling to answer this suitably!? Do you have nay suggestions?

I have no idea what your difficulty is here.

 

[AN630078]

I have no idea what your difficulty is here.

What is the difference in gravitational potential between the top and bottom of the hill?

Is it just 7357.5 J - 0J = 7357.5 J ?

If so how could i elaborate upon this?

 

[PeroK]

What is the difference in gravitational potential between the top and bottom of the hill?

Is it just 7357.5 J - 0J = 7357.5 J ?

If so how could i elaborate upon this?

Why do you need to elaborate on it? The difference is $mgh$.

 

[PeroK]

The Moon has a gravitational field near its surface one sixth of that of the Earth. Calculate:
1. The weight of a man of mass 90 kg on the Moon.
2. The gravitational potential energy he would gain in climbing a 50 m hill on the Moon.
3. The difference in gravitational potential between the top and bottom of the hill.

Okay, I think I see the point of question 3). It's asking for the difference in gravitational potential (which is potential energy per unit mass). It's not asking for the potential energy you found in part 2).

Apologies, I should have noticed that.

 

[AN630078]

Okay, I think I see the point of question 3). It's asking for the difference in gravitational potential (which is potential energy per unit mass). It's not asking for the potential energy you found in part 2).

Apologies, I should have noticed that.

Thank you for your reply. Yes, I think this is what I was trying to say in my original post but was not sure how, which is why I used the formula Vgrav=-GM/r. In which case, how does that affect answering question 3?

 

[PeroK]

Thank you for your reply. Yes, I think this is what I was trying to say in my original post but was not sure how, which is why I used the formula Vgrav=-GM/r. In which case, how does that affect answering question 3?

Okay, that was my fault. First, you should understand why we use $mgh$ instead of $-GMm/r$. We assume the large object has a radius of $R$, so that the gravitational potential at the surface is $V = -GM/R$.
The potential at a height $h$ about the surface is $V(h) = -GM/(R+h)$.

Now, we can take the difference: $$\Delta V = -\frac{GM}{R+h} - ( -\frac{GM}{R}) = GM\frac{h}{R(R+h)}$$ And if $h \ll R$ then $R + h \approx R$ this simplifies to: $$\Delta V \approx \frac{GM}{R^2}h = gh$$ where $g$ is the magnitude of the surface gravity.

Finally, if we reset the zero point of the potential to be the surface, then we have $V = gh$ for $h \ll R$. Or, in terms of the potential energy of a mass $m$ near the surface: $PE = mgh$.

 

[AN630078]

Okay, that was my fault. First, you should understand why we use $mgh$ instead of $-GMm/r$. We assume the large object has a radius of $R$, so that the gravitational potential at the surface is $V = -GM/R$.
The potential at a height $h$ about the surface is $V(h) = -GM/(R+h)$.

Now, we can take the difference: $$\Delta V = -\frac{GM}{R+h} - ( -\frac{GM}{R}) = GM\frac{h}{R(R+h)}$$ And if $h \ll R$ then $R + h \approx R$ this simplifies to: $$\Delta V \approx \frac{GM}{R^2}h = gh$$ where $g$ is the magnitude of the surface gravity.

Finally, if we reset the zero point of the potential to be the surface, then we have $V = gh$ for $h \ll R$. Or, in terms of the potential energy of a mass $m$ near the surface: $PE = mgh$.

Thank you very much for your reply, so would the gravitaional potential be equal to ∆V=gh which would be 1.635 *50 =81.75 Jkg^-1 at the top of the hill?

And would the height at the surface of the Moon be 0m, so ∆V=1.635 * 0 = 0 Jkg^-1

So the difference in gravitational potential would be 81.75-0=81.75 Jkg^-1 ?

 

[AN630078]

Okay, that was my fault. First, you should understand why we use $mgh$ instead of $-GMm/r$. We assume the large object has a radius of $R$, so that the gravitational potential at the surface is $V = -GM/R$.
The potential at a height $h$ about the surface is $V(h) = -GM/(R+h)$.

Now, we can take the difference: $$\Delta V = -\frac{GM}{R+h} - ( -\frac{GM}{R}) = GM\frac{h}{R(R+h)}$$ And if $h \ll R$ then $R + h \approx R$ this simplifies to: $$\Delta V \approx \frac{GM}{R^2}h = gh$$ where $g$ is the magnitude of the surface gravity.

Finally, if we reset the zero point of the potential to be the surface, then we have $V = gh$ for $h \ll R$. Or, in terms of the potential energy of a mass $m$ near the surface: $PE = mgh$.

Thank you for your help, can I ask would my solutions to parts 1 and 2 be correct or do I need to amend these also?

 

[jbriggs444]

Thank you for your help, can I ask would my solutions to parts 1 and 2 be correct or do I need to amend these also?

You are fine. @PeroK is just going through the mathematical underpinnings of the fact that the two formulas for gravitational potential difference: $PE=g(h_1 - h_0)$ and $PE=-GM(\frac{1}{r_1}-\frac{1}{r_0})$ are excellent approximations to each other as long as $h=r_1-r_0$ is much smaller than $r_0$.

50 meters is quite small compared to the radius of the moon. So the approximation is quite acceptable.

It would be painful to deal with numbers like $r_1$ = 1737150 meters and $r_0$ = 1737100 meters and work out the potential difference using the technically correct formula.

[Also, a bit unsettling since the inputs ($r_1$, $r_0$, $G$ or $M$) are not known precisely enough to make the before and after potentials knowable to the requisite accuracy. Fortunately, one does not need the before and after potentials to be precisely known. It is enough that their difference can be precisely known]