Gravitational Field Strength at the equator and poles

[AN630078]

Homework Statement

Hello, I have found several questions concerning the gravitational field strength of Earth and the equator. I thought them to be rather peculiar since I had not come across any line of questioning similar to this before, therefore, I am uncertain of the methods i have used particularly for question 3. Could anyone comment whether I could improve upon my workings here or offer some further guidance?

Between the poles of Earth and its equator the gravitational field strength varies since the Earth is not perfectly spherical. The measured value of g (and apparent weight) also vary because the Earth is spinning.

1. Taking the radius of the Earth to be 6.4 x 10^6 m, calculate the centripetal force on a 1 kg mass on the equator. (The Earth spins a complete turn in 1 day).
2. The measured value of g at the equator is 9.78, how much would it be if the Earth was not spinning?
3.Using your previous answer for g at the equator, and taking g=9.83 the poles, M = 6.0 x 10^24 kg, and using g = GM/R^2, find out how much further away from the centre of the Earth the equator is than the poles.

Relevant Equations

F= mv^2/r
g = GM/R^2
V=2πr/T

1. The centripetal force is equal to F= mv^2/r.
The velocity of the Earth can be found by:
V=2πr/T
T=1 day = 24 hr*60min*60sec=86400 s
v=2π*6.4 x 10^6/86400 s
v=465.4211 ... ~465 ms^-1 to 3.s.f
Therefore, F=1*465/6.4 x 10^6
F=98/1280000=7.265626 *10^-5 ~7.3 *10^-5 N

Would this be correct since it seems incredibly small, although I suppose this is because it is the force acting on a 1kg mass.

2. There would be no change in gravitational field strength since g depends only on the mass of the Earth and our distance from it since g=GM/r^2. Therefore, the measured value at the equator would still be 9.78 N kg^-1. However, there will be a very slight change in the net downward force acting on an object to hold it to the Earth caused by the centripetal acceleration towards the planet's centre by the Earth's rotation. This would disappear if the Earth was no longer rotating.

a=ω2r
a=(2.3*10^−5)2⋅6.37*10^6=3.4*10^−3 ms−2
Which can essentially be neglected as it is on the order of magnitude 3,000 times smaller than g.

3. Would I rearrange g=GM/r^2 in terms of r?
Multiply both sides by r^2: gr^2=GM
Divide both sides by g: r^2 =GM/g
Square root both sides; r= √GM/g

At the equator;
r= √6.67*10^-11*6.0 x 10^24/9.78
r~√4.09*10^13
r=6396893.418 m

At the poles;
r= √6.67*10^-11*6.0 x 10^24/9.83
r~√4.07*10^13
r=6380603.874 m

Difference is distance from centre of the Earth = Distance at equator - distance at poles
6396893.418 - 6380603.874=16289.544 ~ 16300 m to 3.s.f.

 

[mfb]

Therefore, F=1*465/6.4 x 10^6
F=98/1280000=7.265626 *10^-5 ~7.3 *10^-5 N

I don't understand how you got from the first line to the second but you didn't square the velocity and you forgot the units. Adding units would have made the mistake obvious. Always work with units.

For (2), the measured value of g is "what the scale shows", i.e. the centrifugal force is included.

(3) can be fixed once you get the right answer for (1) and (2).

 

[AN630078]

I don't understand how you got from the first line to the second but you didn't square the velocity and you forgot the units. Adding units would have made the mistake obvious. Always work with units.

For (2), the measured value of g is "what the scale shows", i.e. the centrifugal force is included.

(3) can be fixed once you get the right answer for (1) and (2).

Thank you for your reply.
1. No you are correct, thank you for spotting that.
F=1kg*465^2ms^-1/6.4 x 10^6m
F=0.033785 ~3.4*10^-2 N

Would this be correct?

2. So should I subtract the centrifugal force from the gravitational field strength:
9.78-3.4*10^−3=9.6966 N kg^-1 (which is such a small difference I took it to be negligible)

3. How should I amend this? By substituting that g=9.6966 N kg^-1 at the equator?

 

[mfb]

0.3% is the right range. It's not a big difference but it is crucial for (3).

9.78-3.4*10^−3=9.6966 N kg^-1

Something went wrong here.

For (3) you are given the mass of Earth so you can calculate the radius at the poles and the equator separately and then subtract. Keep enough digits in intermediate results.

 

[haruspex]

calculate the radius at the poles and the equator separately and then subtract. Keep enough digits in intermediate results.

Or @AN630078 can use the bit of algebra I posted in the satellite thread.

 

[AN630078]

0.3% is the right range. It's not a big difference but it is crucial for (3).Something went wrong here.

For (3) you are given the mass of Earth so you can calculate the radius at the poles and the equator separately and then subtract. Keep enough digits in intermediate results.

Thank you for your reply. Sorry I am a little confused, how should I go about amending my answer to part 2 in order to solve 3?

 

[AN630078]

Or @AN630078 can use the bit of algebra I posted in the satellite thread.

Thank you for your reply. How would that be applicable here if in question 3 we are trying to find the distance between centre of the Earrth and the equator and the distance between the centre and the poles? Using the formula you previously stated would these not have to be known already?

 

[PeroK]

Thank you for your reply. How would that be applicable here if in question 3 we are trying to find the distance between centre of the Earrth and the equator and the distance between the centre and the poles? Using the formula you previously stated would these not have to be known already?

I think numbers are your enemy here. As an example, for part one you had:

Relevant Equations:: F= mv^2/r
g = GM/R^2
V=2πr/T

Then you could simply have expressed the force in terms of the period by substituting the velocity:
$$F = \frac{mv^2}{R} = \frac{m(2\pi R/T)^2}{R} = \frac{4m\pi^2 R}{T^2}$$
That gives you one calculation that you can do directly on a calculator. Instead you went into a long intermediate calculation of the velocity with all the opportunity to go wrong.

For part 2), would the measured force at the equator be greater or less if the Earth were not spinning?

 

[AN630078]

I think numbers are your enemy here. As an example, for part one you had:Then you could simply have expressed the force in terms of the period by substituting the velocity:
$$F = \frac{mv^2}{R} = \frac{m(2\pi R/T)^2}{R} = \frac{4m\pi^2 R}{T^2}$$
That gives you one calculation that you can do directly on a calculator. Instead you went into a long intermediate calculation of the velocity with all the opportunity to go wrong.

For part 2), would the measured force at the equator be greater or less if the Earth were not spinning?

Thank you for your reply and yes I would have to agree. Would the measured force be less if the Earth were not spininning, since the centrifugal force is included in the value of g which would disappear if the Earth were not rotating?

 

[PeroK]

Thank you for your reply and yes I would have to agree. Would the measured force be less if the Earth were not spininning, since the centrifugal force is included in the value of g which would disappear if the Earth were not rotating?

If the Earth started spinning faster what would happen (to effective gravity) at the equator? The centrifugal force is not real. The centripetal force is real.

 

[AN630078]

If the Earth started spinning faster what would happen (to effective gravity) at the equator? The centrifugal force is not real. The centripetal force is real.

If the Earth were to start spinning faster the value of g would decrease, since centrifugal forces act against gravity and are increased if the rotation of Earth increases?

 

[PeroK]

If the Earth were to start spinning faster the value of g would decrease, since centrifugal forces act against gravity and are increased if the rotation of Earth increases?

Yes, so if the Earth stops spinning you get the maximum effective gravitational force, with no force used for centripetal acceleration.

 

[AN630078]

Yes, so if the Earth stops spinning you get the maximum effective gravitational force, with no force used for centripetal acceleration.

Right ok. So does this mean that the value of g at the equator if the Earth stopped spinning would be 9.78 + value of the centrigual force, since this is no longer acting against g?

 

[PeroK]

Right ok. So does this mean that the value of g at the equator if the Earth stopped spinning would be 9.78 + value of the centrigual force, since this is no longer acting against g?

That's what increase means!

For part 3), think about how $g$ varies with $R$ for the same mass. Use $g_e$ and $g_p$ for gravity at the Equator and Poles; and $R_e$ and $R_g$.

Hint: you don't need the numerical value for $G$ or $M$.

 

[AN630078]

That's what increase means!

For part 3), think about how $g$ varies with $R$ for the same mass. Use $g_e$ and $g_p$ for gravity at the Equator and Poles; and $R_e$ and $R_g$.

Hint: you don't need the numerical value for $G$ or $M$.

Well the gravitational field strength is inversely proportional to the square of the distance from the centre of the Earth, g∝1/r^2
So, g e=1/Re^2 and g p=1/Rg^2
So the greater the distance from the centre of the Earth the lesser the strength of the gravitational field.

 

[PeroK]

Well the gravitational field strength is inversely proportional to the square of the distance from the centre of the Earth, g∝1/r^2
So, g e=1/Re^2 and g p=1/Rg^2
So the greater the distance from the centre of the Earth the lesser the strength of the gravitational field.

What is $R_e/R_g$?

 

[AN630078]

What is $R_e/R_g$?

Sorry but I do not know?

 

[PeroK]

Sorry but I do not know?

It's not supposed to be something you know, it's supposed to be something you can calculate!

 

[PeroK]

If this is too complicated, then you can always calculate $R_e$ and $R_p$ from $G, M$ and $g_e, g_p$ and subtract one from the other.

 

[AN630078]

If this is too complicated, then you can always calculate Re and Rp from G,M and ge,gp and subtract one from the other.

I am sorry but I am confused. How would I calculate R e and Rp from G,M and ge, gp? Would this be using g=GM/r^2 as I did originally? I have still not found the value of g at the equator though? If this is 9.78 + value of the centrigual force would this be;
9.78 + F where F=3.4*10^-2 N as calculated in part 1.
So, 9.81+3.4*10^-2=9.814 N kg^-1

Then r= √GM/g

At the equator;
r= √6.67*10^-11*6.0 x 10^24/9.814
r=6385802.982m

At the poles;
r= √6.67*10^-11*6.0 x 10^24/9.83
r=6380603.874 m

Difference in distance from centre of the Earth = 6385802.982 - 6380603.874=5199.107838~ 5200 m to 3.s.f.

Would this be correct?

 

[mfb]

You need to pay more attention when writing down calculations

9.78 + F where F=3.4*10^-2 N as calculated in part 1.

That part is okay.

So, 9.81+3.4*10^-2=9.814 N kg^-1

What is that? The right side is the right result from the calculation a line above but where does 9.81 come from on the left side?

Difference in distance from centre of the Earth = 6385802.982 - 6380603.874=5199.107838~ 5200 m to 3.s.f.

I wouldn't give three significant figures - two at most, but in practice only the first one has any meaning.
It's not matching the actual difference Earth has because the number given for the equator doesn't fit, but that's a problem of the homework question not a problem of the answer.

 

[AN630078]

You need to pay more attention when writing down calculationsThat part is okay.What is that? The right side is the right result from the calculation a line above but where does 9.81 come from on the left side?I wouldn't give three significant figures - two at most, but in practice only the first one has any meaning.
It's not matching the actual difference Earth has because the number given for the equator doesn't fit, but that's a problem of the homework question not a problem of the answer.

Thank you for your reply. Sorry, I mistook the value fo g at the equator when typing! I meant g=9.78+3.4*10^-2=9.814 N kg^-1.

Thank you for your suggestion, would it be preferable to give the distance as 5000m then? Also, you mention this does not match the actual difference Earth has, however, in the context of the information given in the question would my solutions now be correct?

 

[mfb]

Yes, 5000 m would be better. Unless I missed another mistake it's the right answer.

 

[AN630078]

Yes, 5000 m would be better. Unless I missed another mistake it's the right answer.

Thank you very much for your help I greatly appreciate it, I would not have been able to understand this problem with it