Calculating Height of a Twice-Daily Orbit Above Earth's Surface

[somekid99]

Homework Statement

The question is:
A top secret spy satellite is designed to orbit the Earth twice each day (i.e, twice as fast as the Earth's rotation). What is the height of this orbit above the Earth's surface?

Circumference of orbit not given. Mass not given. Centripetal force is not given.

All I probably know that g is 9.8m/s²

Homework Equations

Fc = m*v²/r
Fg = Gm₁m₂/r²
Fc = m*g
v = 2 pi r/T
T = 12 hours = 43200 seconds
G = 6.674E-11

The Attempt at a Solution

r = (Fc/m)/v²
But I don't know Fc, m, or v.

I tried everything (like substituting v for 2 pi r/T but my teacher keeps saying that there's a way. I feel that there is not enough information given. Maybe someone can find something that I'm not seeing.

 

[Orodruin]

All I probably know that g is 9.8m/s2

No you don’t. This is only true on the Earth’s surface. You can however use it to find an expression for some combination of constants.

Consider what force must be the centripetal force. There is definitely enough information.

 

[somekid99]

No you don’t. This is only true on the Earth’s surface. You can however use it to find an expression for some combination of constants.

Consider what force must be the centripetal force. There is definitely enough information.

Centripetal force would be the same force as Fg but there is no way to know since the only thing I know for certain is that the period is 12 hours.

 

[Orodruin]

Centripetal force would be the same force as Fg but there is no way to know since the only thing I know for certain is that the period is 12 hours.

What do you mean there is no way to know? You need to work a bit with several relations. What do you get if you assume that the gravitational force is the centripetal force?

 

[somekid99]

What do you mean there is no way to know? You need to work a bit with several relations. What do you get if you assume that the gravitational force is the centripetal force?

What do you mean? I still get nothing if I assume that gravitational force is the centripetal force because I don't know the masses of Earth and the satellite. There is no way to know because the only information I have is the period being 12 hours.

The only equation I can think of using period is:
v = 2pir/T

I can try to substitute that into r = (Fc/m)/v²

r = (Fc/m)/(2pir/T)²

Still not sure how I can find r from this.

 

[Orodruin]

What do you mean? I still get nothing if I assume that gravitational force is the centripetal force because I don't know the masses of Earth and the satellite.

You are wrong and if you are not going to accept guidance this thread is just an exercise in futility. It would be more constructive if you instead try following the hints you are given and show what you get. Then we can work on resolving any problems and obstacles you encounter.

 

[somekid99]

You are wrong and if you are not going to accept guidance this thread is just an exercise in futility. It would be more constructive if you instead try following the hints you are given and show what you get. Then we can work on resolving any problems and obstacles you encounter.

I'm sorry but I really don't understand what hints your giving.

I don't understand, what do you mean by "What do you get if you assume that the gravitational force is the centripetal force?"
If I assume that gravitational force is the centripetal force then I get the centripetal force? I don't get what you are saying.

You also said that I should follow the hints, I don't get it. What is there to follow from knowing Fg?

You asked me "what did I meant that there no way to know?" I don't know how to find Fg, Fc, masses, and velocity all from only knowing the period. I think that this question is unsolvable, I really don't know how else to solve it given the information.

 

[Orodruin]

I think that this question is unsolvable

And I am telling you that it is not. Giving up will not solve it however.

I believe you need to work on not assuming that you need to be able to find a number at every single step. Several of the factors that you have mentioned actually do not enter into the final result at all.

I don't understand, what do you mean by "What do you get if you assume that the gravitational force is the centripetal force?"
If I assume that gravitational force is the centripetal force then I get the centripetal force? I don't get what you are saying.

You have an expression for the centripetal force and you have an expression relating the centripetal force to the velocity. Insert the expression for the centripetal force into the relation relating it to the velocity!

Either way, you need to do the actual work and see what you might be missing. Do it step by step and when you get stuck we can discuss how you could find an expression for what you are searching for.

 

[somekid99]

And I am telling you that it is not. Giving up will not solve it however.

I believe you need to work on not assuming that you need to be able to find a number at every single step. Several of the factors that you have mentioned actually do not enter into the final result at all.You have an expression for the centripetal force and you have an expression relating the centripetal force to the velocity. Insert the expression for the centripetal force into the relation relating it to the velocity!

Okay,

So v = 2pir/T
and Fc = m*v²/r

Fc = m* (2pir/T)/r

or if solving for r

r = (2pir/T) / (Fc/m)

 

[Orodruin]

No, you are still not using the fact that the gravitational force needs to be equal to the centripetal force.

Edit: You also forgot to square the velocity and that the velocity contains an $r$.

 

[somekid99]

No, you are still not using the fact that the gravitational force needs to be equal to the centripetal force.

Edit: You also forgot to square the velocity and that the velocity contains an $r$.

Right,

Fc = m* (2pir/T)²/r

r = (2pir/T)² / (Fc/m)

How do I use Fg in this?

Fg = Gm₁m₂/r²

m* (2pir/T)²/r = Gm₁m₂/r²

 

[Orodruin]

So now you have three masses in your expression, what are those masses?

 

[somekid99]

So now you have three masses in your expression, what are those masses?

m = satellite's mass,
m₁/m₂= either Earth's or satellite's mass

 

[Orodruin]

So you have two separate symbols for the same quantity (the satellite's mass). That is generally a bad idea so what happens when you instead use the same symbol to represent the satellite's mass everywhere? (I would suggest using $m$ for the satellite's mass and $M$ for the Earth's mass to keep things clean.)

 

[somekid99]

So you have two separate symbols for the same quantity (the satellite's mass). That is generally a bad idea so what happens when you instead use the same symbol to represent the satellite's mass everywhere? (I would suggest using $m$ for the satellite's mass and $M$ for the Earth's mass to keep things clean.)

m* (2pir/T)²/r = GMm/r²

so like this?

 

[Orodruin]

Yes. So can you solve for $r$ from that equation?

 

[somekid99]

Yes. So can you solve for $r$ from that equation?

m* (2pir/T)²/r = GMm/r²

r = m * (2pir/T)²/(GMn/r²)

I'm not sure if I can simplify the equation.

 

[Orodruin]

No, you have not solved for $r$, it appears on both sides of the equation. Furthermore, there is a particular simplification you can do (it has something to do with a quantity that appears in both the numerator and denominator).

 

[somekid99]

No, you have not solved for $r$, it appears on both sides of the equation. Furthermore, there is a particular simplification you can do (it has something to do with a quantity that appears in both the numerator and denominator).

Sorry for the late response, I had to relearn how to do my algebra hehe.

m*(2pir/T)²/r= GMm/r²
m*(4pi²r²/T²)/r = GMm/r²
4pi²r²m/rT²=GMm/r²
4pi²rm/T²=GMm/r²
4pi²rm*r²=GMm*T²
4pi²r³m=GMnT²
r³=GMmT²/4pi²m
r³=GMT²/4pi²
r= cuberoot(GMT²/4pi²)

All I need now is Earth's Mass.

Wait isn't this Kepler's third law?

 

[PeroK]

Sorry for the late response, I had to relearn how to do my algebra hehe.

m*(2pir/T)²/r= GMm/r²
m*(4pi²r²/T²)/r = GMm/r²
4pi²r²m/rT²=GMm/r²
4pi²rm/T²=GMm/r²
4pi²rm*r²=GMm*T²
4pi²r³m=GMnT²
r³=GMmT²/4pi²m
r³=GMT²/4pi²
r= cuberoot(GMT²/4pi²)

All I need now is Earth's Mass.

Wait isn't this Kepler's third law?

That looks good and, yes, that is a special case of Kepler's third law for a spherical orbit.

Note that when you don't know the numeric value of a variable in a question (such as the mass of the satellite), you should just go ahead with a vairable ($m$ in this case) and see what happens. The physics and the maths working is essentially the same. If, as happened in this case, the variable cancels out of the equation, then you are fine. If not, then you do have a problem.

In this case, the mass of the Earth did not cancel out, so you need to look that up. This means that a satellite that orbits Mars, say, with the same period would have a different orbital radius. So, by leaving the Earth's mass as $M$ (rather than plugging in a horrible number at the start) you learned this as well - and correctly identified the Kepler's law.

It's a good example of why algebra is vastly superior and more powerful than arithmetic. Keeping things as variables like $m$, $M$ and $r$ let's you see the physics emerge from the algebra.

 

[haruspex]

All I need now is Earth's Mass.

Not needed if you know Earth's radius, R, and g there. Write an expression for g in terms of G, M and R.

 

[PeroK]

Not needed if you know Earth's radius, R, and g there. Write an expression for g in terms of G, M and R.

I imagine most people would have to look up the Earth's radius in any case!

 

[Orodruin]

I imagine most people would have to look up the Earth's radius in any case!

Well, it is one less variable to look up (in the case of the mass you also need to look up the gravitational constant). Furthermore, in putting the expression in terms of $g$ at the surface, the result can be expressed in multiples of the Earth radius directly, which will not give you a numerical value, but still give you some sort of relation to a length.

@somekid99 Also note that the question asks for the height above the surface, not the radius at which the satellite is orbiting. These are not the same quantities unless the Earth's radius is negligible in comparison. Just think about this before you give your final answer.

 

[somekid99]

Well, it is one less variable to look up (in the case of the mass you also need to look up the gravitational constant). Furthermore, in putting the expression in terms of $g$ at the surface, the result can be expressed in multiples of the Earth radius directly, which will not give you a numerical value, but still give you some sort of relation to a length.

@somekid99 Also note that the question asks for the height above the surface, not the radius at which the satellite is orbiting. These are not the same quantities unless the Earth's radius is negligible in comparison. Just think about this before you give your final answer.

If I plug in, the gravitational constant and Earth's mass from the previous questions in my book,

G = 6.67408E-11
M = 5.972E24
T = 43200 s

r = cuberoot(6.67408E-11] * 5.972E24 * 43200²/4pi²
r = 26609680.17 m

But this is from the center of the Earth not from the surface.

I know myself that the distance from the core to surface is around 6000km but I can't prove it.

 

[Orodruin]

G = 6.67408E-11
M = 5.972E24

You are missing units here. Even if they are the SI units, you should always write them out.

To be honest, I would follow the suggestion to work out GM in terms of the Earth's radius and g at the surface instead. This will lead you to a result that is a multiple of the Earht's radius and it is very easy to subtract the Earth's radius from it. You can find the Earth's radius in many places, for example https://en.wikipedia.org/wiki/Earth

r = 26609680.17 m

This number has way too many significant digits. Never give more precision than what is reasonable to give.

 

[somekid99]

You are missing units here. Even if they are the SI units, you should always write them out.

To be honest, I would follow the suggestion to work out GM in terms of the Earth's radius and g at the surface instead. This will lead you to a result that is a multiple of the Earht's radius and it is very easy to subtract the Earth's radius from it. You can find the Earth's radius in many places, for example https://en.wikipedia.org/wiki/EarthThis number has way too many significant digits. Never give more precision than what is reasonable to give.

G = 6.67408E-11 m³kg^-1s^-2
M = 5.972E24 kg
oops I forgot about sigfigs.
r = 266609680 m.

I don't understand, how do I use Earth's radius and g at the surface for G and M? Isn't M the mass the of the Earth?

 

[Orodruin]

G = 6.67408E-11 m3kg-1s-2
M = 5.972E24 kg

Better.

oops I forgot about sigfigs.
r = 266609680 m.

Still far too many. As a rule of thumb when multiplying numbers, the result should not have more significant digits than the number with the fewest significant digits that you put in. (Keep all decimals in intermediate steps though.)

I don't understand, how do I use Earth's radius and g at the surface for G and M? Isn't M the mass the of the Earth?

Yes, but knowing that g is 9.8 m/s you can get rid of the need to look this up at all. What is the gravitational force on an object at the Earth’s surface? (Both in terms of Newton’s law of gravitation and in terms of the gravitational acceleration g at the surface.)

 

[somekid99]

Better.
Still far too many. As a rule of thumb when multiplying numbers, the result should not have more significant digits than the number with the fewest significant digits that you put in. (Keep all decimals in intermediate steps though.)
Yes, but knowing that g is 9.8 m/s you can get rid of the need to look this up at all. What is the gravitational force on an object at the Earth’s surface? (Both in terms of Newton’s law of gravitation and in terms of the gravitational acceleration g at the surface.)

So how do you write the significant figure then? If a number is in scientific notation do I follow the actual number or just use the same amount of digits as the scientific notation? So if the one with the least digits, is 5.972 E24 I can only write 4 digits?

at Earth's surface the gravitational force so:

Fg = m*g
mg = GMm/r²
mgr² = GMm
mgr²/Gm= M
gr²/G = M

From the wiki, the radius of Earth is 6371.0 km = 6371000 m
and gravity at the surface is 9.8m/s²
G is 6.67408E-11 m³kg^-1s^-2
9.8 * 6371000²/6.67408E-11 = M
5.960049652E24 = M ?
significant figures: 5.9E24 kg = M?

 

[Orodruin]

mg = GMm/r2

You have the correct relation, but you are not solving for the correct thing. As was noted by PeroK on the previous page, you should keep the algebraic expressions as long as possible rather than trying to find numerical values on the way. This is what you are after:

gr2/G = M

but it is not true for any $r$, just for the Earth's radius. You should be careful not to use the same symbol for different quantities just as you should be careful not to introduce several different notations for the same quantity. I suggest calling the Earth's radius R, which would give you $M = gR^2/G$. What happens if you insert this into the expression for your sought radius?

 

[somekid99]

You have the correct relation, but you are not solving for the correct thing. As was noted by PeroK on the previous page, you should keep the algebraic expressions as long as possible rather than trying to find numerical values on the way. This is what you are after:

but it is not true for any $r$, just for the Earth's radius. You should be careful not to use the same symbol for different quantities just as you should be careful not to introduce several different notations for the same quantity. I suggest calling the Earth's radius R, which would give you $M = gR^2/G$. What happens if you insert this into the expression for your sought radius?

Wait wait, I'm a bit confused.
So M = gR²/G
R = squareroot(MG/g)

sqrt(MG/g) = cbrt(GMT²/4pi²)

 

[haruspex]

Wait wait, I'm a bit confused.
So M = gR²/G
R = squareroot(MG/g)

sqrt(MG/g) = cbrt(GMT²/4pi²)

You are confusing two radii. On the left is R, the radius of the Earth; on the right is r, the orbital radius of your satellite.
Use R = squareroot(MG/g) to get rid of the MG factor in your expression for r.

 

[somekid99]

You are confusing two radii. On the left is R, the radius of the Earth; on the right is r, the orbital radius of your satellite.
Use R = squareroot(MG/g) to get rid of the MG factor in your expression for r.

So if R is the radius of Earth and r is the radius of the the satellite's orbit then to find the height of the satellite:
Height = cbrt(GMT²/4pi²) - sqrt(MG/g)

Or what do you mean by get rid of the MG factor?
So I substitue MG with R?
r = cbrt((sqrt(MG/g))T²/4pi²)?

 

[Orodruin]

So if R is the radius of Earth and r is the radius of the the satellite's orbit then to find the height of the satellite:
Height = cbrt(GMT2/4pi2) - sqrt(MG/g)

You were supposed to replace GM with gR^2, not the other way around. All that happened now was that you complicated the expression. The point was to get rid of constants you did not know.

 

[somekid99]

You were supposed to replace GM with gR^2, not the other way around. All that happened now was that you complicated the expression. The point was to get rid of constants you did not know.

ohhh, I thought that was to prove what the radius of the Earth was, that's why I got confused.

r = cbrt(gR²T²/4pi²)
r = cbrt(9.8*6371000²43200²/4pi²)
r = 26591919.1 m

Then I subtract r with Earth's radius
26591919.1 m - 6371000 m = 20220919.1 m

and that's the height.

 

[Orodruin]

Apart from the fact that you again have way too many significant digits. You really do not have more than two at best. I suggest writing the answer in units of Mm (megameter) or, even in units of the Earth's radius. Giving nine significant digits you are essentially saying you know all the input variables to a precision that would be the same as knowing your height with a precision of the size of an atom. This is completely unreasonable.