# A few exam questions

I just got back from doing my physics exam, didn't do so well, but I did well during the semester and I think I did well on a final summative project.
This is gr. 11 by the way, so nothing to fancy.

Anyway, I wrote down the questions i wasn't sure of, and I just want someone to help me out with the answers please:

1.

4 students timed(something) and got the times 1.85, 1.8, 1.9, and 2, what was the average?

I know this looks really basic and all, the answer I got was 2, simply because of following the significant figure rules:

1.85 + 1.8 + 1.9 + 2 = 7.55, accurate to no dec. place because of the number '2'.

Therefore it's 8, divide by 4 = 2, accurate to one sig fig, wierd result yes, but that's how I did it, look alright?

2.

Which of the following demonstrates newton's 3rd law?

turning a corner
2 skaters pushing against one another and being sent backwards
earth held in orbit by the sun
friction of driving on a road(I think)
car tires slipping

I think it's the one with skaters, although in some sense the 3rd law can be seen anywhere, I also thought the earth one was possible, because the sun pulls the earth, and so the earth pulls the sun, I don't know, someone please tell me if I'm right, it wasn't the best question I know but...

3.

What is true about the gravitational field strength of earth?

9.8 everywhere
greatest at equator
smallest at the top of mt. everest
largest at the deepest spot in the ocean
largest at the poles

It's one of the last two I'm pretty sure, though really only the last makes sense, sort of since the earth bulges at the equator, and the radius is smaller at the poles. Though none are really true, because you can go higher than mount everest, and lower than the ocean floor and the poles, but what would you say?

4.

Which of the following does not contribute to kinetic friction?

mass
coefficient of friction
normal force
applied force
gravitaitional field strength

I guessing normal force, though I'm not sure; I was absent when we did most friction related stuff, could someone correct and explain why?

5.

What's the angle of the reflected ray with an incident ray of 34 degrees?

0
34
56
66
74

I'm thinking 56, because we were told that reflected rays make a 90 degree angle with the incident ray... so 90 - 56 = 34, am I right?

6.

What's the critical angle of ethyl alchohol(index of refraction 1.36) in air(index of refraction 1.0003)? Or was it the other way around?

47.3
0.0257
90.0
48.8
45.0

I'm thinking 47.3

sin(critical angle) = n_2/n_1
theta = 1.0003/1.36 sin(inverse)
= 47.35 or something like that, am I right?

Hootenanny
Staff Emeritus
Gold Member
Byrgg said:
4 students timed(something) and got the times 1.85, 1.8, 1.9, and 2, what was the average?

I know this looks really basic and all, the answer I got was 2, simply because of following the significant figure rules:

1.85 + 1.8 + 1.9 + 2 = 7.55, accurate to no dec. place because of the number '2'.

Therefore it's 8, divide by 4 = 2, accurate to one sig fig, wierd result yes, but that's how I did it, look alright?
I'm afraid not, you should only round at the end of the calculation; not mid calculation. Therefore, you should obtain;

$$\frac{7.55}{4} = 1.8875 = 1.89 \; (\text{3sf})$$
Byrgg said:
Which of the following demonstrates newton's 3rd law?

turning a corner
2 skaters pushing against one another and being sent backwards
earth held in orbit by the sun
friction of driving on a road(I think)

car tires slipping

I think it's the one with skaters, although in some sense the 3rd law can be seen anywhere, I also thought the earth one was possible, because the sun pulls the earth, and so the earth pulls the sun, I don't know, someone please tell me if I'm right, it wasn't the best question I know but...
All the red ones are examples of newton's third law. You can think of friction being the force the tyres exert on the earth, if the earth didn't push back the wheels would just spin and you wouldn't go anywhere. EDIT: Actually, they're all examples of newton's thrid law as you say, as newton's third law is valid everywhere. But the example were it would be most obvious is the skaters as you correctly said.

Byrggs said:
What is true about the gravitational field strength of earth?

9.8 everywhere
greatest at equator
smallest at the top of mt. everest
largest at the deepest spot in the ocean
largest at the poles

It's one of the last two I'm pretty sure, though really only the last makes sense, sort of since the earth bulges at the equator, and the radius is smaller at the poles. Though none are really true, because you can go higher than mount everest, and lower than the ocean floor and the poles, but what would you say?
I would agree with you there. Besides, once you go below the surface of the earth (including any water), Newton's law of gravitation breaks down.
Byrggs said:
Which of the following does not contribute to kinetic friction?

mass
coefficient of friction
normal force
applied force
gravitaitional field strength

I guessing normal force, though I'm not sure; I was absent when we did most friction related stuff, could someone correct and explain why?
I'm afraid your guess is wrong by a country mile. What is the equation forthe kinetic frictional force?
Byrggs said:
What's the angle of the reflected ray with an incident ray of 34 degrees?

0
34
56
66
74

I'm thinking 56, because we were told that reflected rays make a 90 degree angle with the incident ray... so 90 - 56 = 34, am I right?
I'm afraid this is also wrong. What is the law of reflection? Probably the first rule you learned when considering mirrors.
Byrggs said:
What's the critical angle of ethyl alchohol(index of refraction 1.36) in air(index of refraction 1.0003)? Or was it the other way around?

47.3
0.0257
90.0
48.8
45.0

I'm thinking 47.3

sin(critical angle) = n_2/n_1
theta = 1.0003/1.36 sin(inverse)
= 47.35 or something like that, am I right?
I would agree with you there. Although none of the given answers are technically correct because 47.35 = 47.4(3sf)

Last edited:
(this is in response to hootenanny, jsut in case someone else posts before I'm done)

Err, yes I considered that for the first one as well, the teacher showed us a similar example where I think the way was like what I did, so really the answer is wrong, I figured, but someone else(with a bit higher average than me) got the same answer, I'll see how my teacher marks it, I agree with you, I was originally going to do that, and of course my first impreesion was right(changing my first impression cost me several more marks on the test too...)

The second one, yes, I think it sounds right, but turning a corner applies this too right? (applying a force outwards I believe, and the road responds by pushing you inwards, though I think there was another bit in that choice... car tires slipping(on an icy surface I'm pretty sure was the other part) maybe as well, well the 3rd law is present anywhere just about. Hmm, if you are applying a force to the ice(backwards) to move, then shouldn't the ice still push you forwards? I think I'm missing something here, could you clarify?

The third one... I don't know, what did you mean by the gravitaion breaking down in the water?

Fourth, ugh like I said, I was absent( I know I should've gotten the notes), I don't know about the equation of which you speak, and I no longer have my textbook to refer too, could you help me out here?

Fifth, I heard the incident ray made a right angle with the reflected ray, what went wrong here?

Last one, yeah, I didn't think anything was right either, that one was just the closest so... yeah.

Oh, here's one more, I'm pretty sure about it, just need to make sure:

A rocket reaches an altitude of 4.50 X 10 ^ 2 m in 4.0s, what is the accleration, what is the velocity?

Here's what my friend showed me(for the first part):

d = v_1(t) + 0.5(a)(t ^ 2)

I was going to use that one too...

v_1 = 0 therefore

d = 0.5a(t^2)
2d =a(t^2)
a = (2d)/(t^2)
= ((2)(450))/4.0^2
= 900/16
= 65.25 m/s^2

Is that right?

The velocity is easy:

v = d/t
= 450/4
= 112.5

Right?

Oh wait, is it the other one?

v_2 = v_1 + t(a)?

Hootenanny
Staff Emeritus
Gold Member
Byrgg said:
(
The second one, yes, I think it sounds right, but turning a corner applies this too right? (applying a force outwards I believe, and the road responds by pushing you inwards, though I think there was another bit in that choice... car tires slipping(on an icy surface I'm pretty sure was the other part) maybe as well, well the 3rd law is present anywhere just about. Hmm, if you are applying a force to the ice(backwards) to move, then shouldn't the ice still push you forwards? I think I'm missing something here, could you clarify?
I edited my post after some consideration (obviously while you were replying)
Byrggs said:
The third one... I don't know, what did you mean by the gravitaion breaking down in the water?
The easiest way to think about this is when you are outside or on the surface of the earth, all the gravitational force created by all the molecules which make up the earth are all pulling in the same direction, towards the centre of gravity. Now, if you go below the surface, such as under water, the water molecules will be pulling you in all different directions, the ones above you will be pulling you up, the ones to either side will be pulling you sideways etc. Do you follow?
Byrggs said:
Fourth, ugh like I said, I was absent( I know I should've gotten the notes), I don't know about the equation of which you speak, and I no longer have my textbook to refer too, could you help me out here?
I'm not happy with this question, becuase all the factors listed can affect the size of the frictional force. Were there no other constraints? For your reference the kinetic frictional force is given by $F = \mu R$, where R is the normal reaction force. This question is poorly worded and impossible to answer. I am dissapointed with your physics teacher if the question was presented this way.
Byrggs said:
Fifth, I heard the incident ray made a right angle with the reflected ray, what went wrong here?
Thats wrong, thats what went wrong. Quite simply, the angle of reflection is equal to the angle of incedence. Where did you here this rule of yours?
Byrggs said:
A rocket reaches an altitude of 4.50 X 10 ^ 2 m in 4.0s, what is the accleration, what is the velocity?

Here's what my friend showed me(for the first part):

d = v_1(t) + 0.5(a)(t ^ 2)

I was going to use that one too...

v_1 = 0 therefore

d = 0.5a(t^2)
2d =a(t^2)
a = (2d)/(t^2)
= ((2)(450))/4.0^2
= 900/16
= 65.25 m/s^2
That's probably a type but the correct answer is 56.25 m.s-2
Byrggs said:
The velocity is easy:

v = d/t
= 450/4
= 112.5

Right?
I'm afraid thats wrong, the above formula assumes constant velocity, which we haven't got in this case. Your second formula would be the one to use;
Byrggs said:
v_2 = v_1 + t(a)?

Woops yeah, the answer was 56.25... Also, what about the car on ice question, I'm a little curious, if the car applies a force backwards, the ice pushes forwards, but as we all know, driving on ice isn't easy, could you explain what else is going on here? I know you editted your post, but I'm just curious...

And yes, I always thought the angle of relfection was equal to the angle of incidence, I'm not sure where the right angle notion came from exactly...

Also, why do the particles in water pull you in different directions? The center of gravity is still beneath you, could you explain please?

Also, I'm not entirely sure about the friction factors one.So the kinetic friction is a result of whatever greek character you used there, and the normal force? I thought mass played a role as well? There was also a free-body diagram with it, looking something like this:

(consider O to be the object, and ignore dots)

..........................|
..........................|(normal)
..........................|
..(friction) ------- O ------- (applied)
..........................|
..........................|(gravity)
..........................|

Don't know if this helps any...

And inside the object was velocity(in direction of applied) and I think the letter you used in that equation(for kinetic friction) was also inside. Also for that question, it might've said that greek letter instead of coefficient of friction for that option, not entirely sure if that helps at all either.

Last edited:

nrqed
Homework Helper
Gold Member
Byrgg said:
4.

Which of the following does not contribute to kinetic friction?

mass
coefficient of friction
normal force
applied force
gravitaitional field strength

I guessing normal force, though I'm not sure; I was absent when we did most friction related stuff, could someone correct and explain why?
.

Well, ALL of them may contribute to the kinetic friction force!! So it is a very badly written question.

The kinetic frinction force has a magnitude $\mu_k N$ where $\mu$ (Greek letter "mu", pronounced "mew") is the coefficient of kinetic friction and N is the magnitude of the normal force.

Depending on the inclination of the surface on which the object and the direction of the applied force (if any), the magnitude of the normal force may or may not depend on the mass, the value of g or the applied force!

EDIT: I just saw your drawing. In *that* case, the normal force is simply mg (the applied force does not affect the normal force) so in *that* case, the answer was "the applied force"
Pat

Ok a few more about that one, the normal force is only dependent on mg in that one right? Like you said, then what it is the coefficient of friction dependent on? I believe I read about it a bit in my textbook, but I don't have it any more since my exam is done, what is it dependent on? Oh, and how do you write the 'mu' as you have, and the subscript? I read the thread about producing latex images, but I couldn't get it working...

Also, how would this change if the object was on an incline as you mentioned?

Oh, and I don't know if you already intended to or not, but would you be able to lend a hand with the other questions I still have, if it's not too much trouble. I appreciate the help you've provided thus far, and I hope you'll continue.

nrqed
Homework Helper
Gold Member
Byrgg said:
Ok a few more about that one, the normal force is only dependent on mg in that one right?
Yes. In fact , the normal force is equal to mg in that case
Like you said, then what it is the coefficient of friction dependent on? I believe I read about it a bit in my textbook, but I don't have it any more since my exam is done, what is it dependent on?
It depends only on the *composition* of the two objects sliding against one another (steel on steel or ice on steel or whatever)
Oh, and how do you write the 'mu' as you have, and the subscript? I read the thread about producing latex images, but I couldn't get it working...
Just go to my post and left click with your mouse on the formula, you will have a small window opening up that will show exactly the code I entered.
Also, how would this change if the object was on an incline as you mentioned?
It depends on how the applied force would be acting. If the applied force would still be parallel to the surface, the friction force would still not depend on the applied force (but the normal would be $mg cos \theta$ instead). However, as soon as the applied force is not entirely parallel to the surface (no matter if the plane is inclined or not), then the normal force will depend on the applied force (its magnitude and the angle at which it is applied).

Oh, and I don't know if you already intended to or not, but would you be able to lend a hand with the other questions I still have, if it's not too much trouble. I appreciate the help you've provided thus far, and I hope you'll continue.
Sure, but I can't reread all the posts so maybe you could tell me which of the questions in your first post are still unanswered?

Ok, so the only things affecting kinetic friction(on a horizontal plane) are the normal force, which is dependent on the force of gravity, so that can be included as well, right? Also, the $\mu_k$, the coefficient of kinetic friction, dependent on the materials in contact. The force of friction in the free-body-diagram was what exactly then? One of the factors going into the calculation of $\mu_k N$?
If the applied force was applied at an angle, then how would that affect the normal force? Would it increase in response to the added force into the ground(and thus providing a greater reaction force parallel to surface). And what about on an inclined plane? How is the normal force affected there, the normal force diminshes as the incline increases right?
If all of this is right, then I can move on to the remaining ones.

Now, about the other questions, the ones still remaining are as follows:

Why does a car slip on ice exactly? If you push the tires backwards into the ice, it should still respond with a reaction force of equal magnitude, right? So what causes the slippping?

Last, why do the molecules pull you in different directions in water, thus affecting gravity apparently?

nrqed
Homework Helper
Gold Member
Byrgg said:
Ok, so the only things affecting kinetic friction(on a horizontal plane) are the normal force, which is dependent on the force of gravity, so that can be included as well, right?
Right. As long as there is no other force with a component perpendicular to the surface. Basically, the normal force is such that $\sum F_y =0$. If there is on ly gravity, you have N-mg=0 so N=mg. If any other force has a y component, the normal force will differ from mg.
Also, the $\mu_k$, the coefficient of kinetic friction, dependent on the materials in contact.
yes
The force of friction in the free-body-diagram was what exactly then? One of the factors going into the calculation of $\mu_k N$?
I am not sure what you mean. The kinetic friction force is not a factor in the calculation of $\mu_k N$, the kinetic friction force *is* $\mu_k N$!

If the applied force was applied at an angle, then how would that affect the normal force?
Yes. (we are still talking about a horizontal plane I assume)
Would it increase in response to the added force into the ground(and thus providing a greater reaction force parallel to surface).
It depends if the applied force has a y component which is negative or positive. In the first case, the applied force is partly "pushing the object into the ground" if you see what I mean. This means that the normal force will be greater than mg and therefore the friction force is larger than when the applied force is horizontal. But you could also have an applied force with a positive y component (think of a string attached to the object and someone pulling on the string at an angle above the horizontal). This will decrease the normal force and therefore decrease the friction force.
And what about on an inclined plane? How is the normal force affected there, the normal force diminshes as the incline increases right?
If all of this is right, then I can move on to the remaining ones.
Again it depends on all the forces in the problem so the question is too vague. Consider the special case of no applied force at all. There is just gravity. Then, as the angle of the incline increases, the component of gravity perpendicular to the surface decreases (it is $mg cos \theta$ with the angle measured with respect to the horizontal)). So you are right that in that case the normal force decreases as the angle increases.
Now, about the other questions, the ones still remaining are as follows:

Why does a car slip on ice exactly? If you push the tires backwards into the ice, it should still respond with a reaction force of equal magnitude, right? So what causes the slippping?
When tires rotate without slipping, the friction force is a static friction force. When the tires are slipping, the friction force is a kinetic friction force (which smaller than the static friction force). But there is still a friction force and there is a reaction force provided by the ground indeed.
But if the surface is very icy, this kinetic friction force is very small.

Imagine being on a perfectly levelled icy surface. You push on the gas pedal. The wheels start spinning. They apply a kinetic friction force on the ground (which is small because the surface is very icy) and the ground apply a reaction force on your car. You *will* start to move albeit with a very small acceleration because the forces are small.

In real life, however, the surface is usually not perfectly levelled and the tires are in some grooves so the force is not enough to overcome gravity keeping you stuck in the grooves and you remain stuck.

Last, why do the molecules pull you in different directions in water, thus affecting gravity apparently?
I am not sure what this means. Of course gravity is unchanged. What is affected is the "apparent weight" but the way this is defined, it is not the true weight which is still mg. The water molecules have random motion but the net force exerted by the water on you is not in "different" directions, there is a net force upward (the buoyant force). So I am a bit confused by this question.

Patrick

Hootenanny
Staff Emeritus
Gold Member
Okay just relating your your question on friction, if we take a particle of mass m moving with a constant velocity with a force is being applied by a light inelastic string at an angle of $\theta$ degrees above the horizontal. The tension in the string is T. Now, as the particle is not acclerating the vector sum of all the forces should be zero. Now the tension will have a component in the horizontal plane ($T\cos\theta$) and a component in the vertical plane ($T\sin\theta$). Now, the friction force is given by $F_{r} = \mu R$, were $\mu$ is the co-efficient of friction and R is the normal reaction force. So if we now sum the forces;

In the horizontal plane we have friction and the applied force;

$$\sum\vec{F_{x}} = T\cos\theta - F_{r} = 0$$

And in the vertical plane we have gravity, the normal force and the applied force;

$$\sum\vec{F_{y}} = R + T\sin\theta - mg = 0$$

Now rearranging the above equation to find a function of R in terms of $\theta$ we have;

$$R = mg - T\sin\theta$$

Now substituting this into our equation for friction we have;

$$F_{r} = \mu R = \mu(mg - T\sin\theta)$$

Can you see now why than angle at which the force is applied affects the normal reaction force and hence the frictional force? Observe that if the rope is horizontal $\theta = 0$ and the reaction force just becomes R = mg. Do you follow?

EDIT: Looks like Patrick got there before me, I need to type faster Hootenanny
Staff Emeritus
Gold Member
Perhaps, the example of water was a bad idea. This is slightly off topic, but I don't think it will do any harm (and I'm not doing anything till later ). The real question is why does newton's law of gravitation break down at the surface of the earth. OKay, so let us take an object of mass m at a point inside the earths surface, a distance r from the centre of the earth thus;
[PLAIN]http://eml.ou.edu/physics/module/newton/jik_leong/src_img/insideearth.jpg [Broken]
Taken from the University of Oklahoma website

An important point to note here is that the matter above the particle exerts no force what so ever on the particle, this is derived from the fact that a uniform shell exerts no gravitational force on an object placed within it. Another point to note is that the gavitational force experienced by the object is only due to the mass of the earth below it, i.e. inside the red ring in the above diagram. As the object moves closer to the centre of the earth, there are two antagonistic effects. The magnitude of the gravitaional force increases due to the reduced distance from the centre of mass (recall that the gravitational field stregnth is inversely proportional to the square of the distance between the two objects. However, the force also decreases due to the decreased mass enclosed by the red circle. If we assume that the earth is of a uniform density the latter is more significant than the former.

As I said above, the mass of the particles above the object do not contribute to the net force and therefore can be ignored. Let the mass of the earth enclosed by the circle be M. The mass enlcosed by the red circle is given by the product of density ($\rho$) and volume (v), assuming the earth is a sphere;

$$M = \rho v = \rho\frac{4}{3}\pi r^{3}$$

Now if we substitute the above into newton's law of gravitation we obtain;

$$F = -\frac{GMm}{r^{2}} = - \frac{Gm}{r^{2}}\cdot\rho\frac{4}{3}\pi r^{3} = -\frac{4\cdot Gm\rho\pi r}{3}$$

Now, if we compare the two equations, you will see that;

$$\frac{4\cdot Gm\rho\pi r}{3} < \frac{GMm}{r^{2}}$$

Provided that r is less than the radius of the earth.

Last edited by a moderator:
nrqed
Homework Helper
Gold Member
Hootenanny said:
Perhaps, the example of water was a bad idea. This is slightly off topic, but I don't think it will do any harm (and I'm not doing anything till later ). The real question is why does newton's law of gravitation break down at the surface of the earth. OKay, so let us take an object of mass m at a point inside the earths surface, a distance r from the centre of the earth thus;
....

I would not say that Newton's law of gravitation is breaking down. It is used in the rest of the calculation. but the equation $G {m_1 m_2 \over r^2}$ is only valid for point masses. What *is* breaking down here is treating the Earth as a point mass, not the law of gravitation itself (and it obviously breaks down here since one could not be inside a point mass:tongue2: ).

Newton's law of gravitation does break down (for example in understanding the precession of the aphelion of the planets, most clearly detectable for Mercury) but not in this example.

I know that Hoot knows all of that, I just wanted to clarify.

Cheers!

Patrick

Hoot, I still don't see why you say that Newton's law would break down. It is indeed universal . The shell theorem is basically a simple application, and if I remember correctly, it was Newton himself who proved it .
$$F = -\frac{GMm}{r^{2}} = - \frac{Gm}{r^{2}}\cdot\rho\frac{4}{3}\pi r^{3} = -\frac{4\cdot Gm\rho\pi r}{3}$$
I think the -ve sign is not required.

Arun

Edit :Patrick's on a roll !

Hootenanny said:
However, the force also increases due to the decreased mass enclosed by the red circle

I thought force was directly proportional to mass, wouldn't the force decrease due decreased mass?

Doc Al
Mentor
Byrgg said:
I thought force was directly proportional to mass, wouldn't the force decrease due decreased mass?
Of course; I'm sure Hoot meant to write that "the force also decreases due to the decreased mass..."

Ok, so how do I explain the original question then? Is the gravitational field strength strongest at the poles, or at the bottom of the ocean, and why exactly? Did I miss something here?

Doc Al
Mentor
Byrgg said:
Ok, so how do I explain the original question then? Is the gravitational field strength strongest at the poles, or at the bottom of the ocean, and why exactly? Did I miss something here?
What point is closest to the Earth's center? (What's the difference between equatorial and polar radii? How deep is the ocean?)

I'm not sure... I never found out anything about that...

Hootenanny
Staff Emeritus
Gold Member
Theres been a fair bit of activity on this thread since my last visit. Yes, the error the Doc Al pointed out was indeed a typo and was duely corrected. With reference to this;
Hootenanny said:
why does newton's law of gravitation break down at the surface of the earth
My diction here was probably not the best. Rather, "why can we no longer treat the earth as a point mass once we are below the surface?" would have been a better idea. It should appear obvious that once we are 'inside' something we consider to be a point mass, we can no longer consider it as such; my post just gives a slightly more objective view.
arunbg said:
I think the -ve sign is not required.
This just indicates that the force is attractive, it save using the vector notations above the r and the F.
Byrgg said:
I'm not sure... I never found out anything about that...
Think about Newton's law of gravition;
$$\vec{F} = \frac{GMm}{\vec{r}^2}$$
The gravitational force is greatest when r is smallest. Where is r (this distance from the centre of the earth to the object) smallest? Ontop of a mountain? At the equator? At the poles? At the bottom of an ocean?

Err, the bottom of the ocean is the smallest radius right? Or is that at the poles... I'm just not really sure...

Hootenanny said:
This just indicates that the force is attractive, it save using the vector notations above the r and the F.
Why can't a positive sign( or no sign at all ) indicate attractive force ?
It's just convention and that is why I said that the -ve sign is not required. Of course you know that, I was simply explaining my comment.

Byrgg, you can google to find the actual values.

Hootenanny
Staff Emeritus
Gold Member
arunbg said:
Why can't a positive sign( or no sign at all ) indicate attractive force ?
It's just convention and that is why I said that the -ve sign is not required. Of course you know that, I was simply explaining my comment.
Its just the way I remember being taught it, of course Newton's law is valid with or without signs.

Ugh, the dreaded google, that's why I had to find a physics help forum in the first place, it's hard sifting through the internet, and I've come to appreciate the help, but I'll go try...

Ok, first let me get this straight, the force is dependent on the mass and the distance between the centers of the 2 objects, I already learned this... Say the radius was smaller at one, this would mean the mass of the earth below you would also decrease, right? Or is this not as relevent because the distance is squared?

I'll be back soon, I've got to go try and find those values...