Energy problem and rotation

[minifhncc]

Homework Statement

A satellite of mass 190kg is placed in orbit 4.5 * 10⁵ m above the surface of the Earth. Find the energy required to put this satellite into this orbit. Ignore air resistance but include Earth's daily rotation in your calculations.

Homework Equations

r^3/T^2 = GM/(4Pi^2)

The Attempt at a Solution

I calculated the change in potiential energy (F=-GmM/R) which yielded 1.68*10¹¹. However I don't think is correct since it doesn't include the rotation.


Thanks

 

[tiny-tim]

Hi minifhncc!

I calculated the change in potential energy …

How about the change in kinetic energy ?

 

[Filip Larsen]

You need to consider the change in both potential and kinetic energy. For change in kinetic energy it seems you need to make some assumptions about where on the Earth the satellite is launched from (for instance, near equator) and into which orbit it is placed (for instance, an east-ward equatorial orbit).

 

[minifhncc]

Hi, the change in kinetic energy is the work done, but the change in gravi. potiential energy is also work done... but obviously they would yield different values. What's the difference between these values? (ie. change in K.E. and change in G.P.E)

Thanks

 

[minifhncc]

Hi again,

I've been trying to get my head around this question. But I don't know how I can include the rotation of the Earth.

Also, what velocity would I use? I worked out the velocity that the satellite should have if it was orbiting at that radius (it was about 7000 m/s, using Kepler's law of periods), but wouldn't that be tangential to the Earth?

Furthermore, how would I consider both the G.P.E. and K.E. ? I really don't get this part.

Thanks again

 

[tiny-tim]

Hi minifhncc!

Hi, the change in kinetic energy is the work done, but the change in gravi. potiential energy is also work done... but obviously they would yield different values. What's the difference between these values? (ie. change in K.E. and change in G.P.E)

Yes, PE is defined as (minus) the work done (by a conservative field, such as gravity).

But how are you going to calcuate the work done in manoeuvring the rocket into orbit?

Just use the definition of KE, not some long short-cut.

… I don't know how I can include the rotation of the Earth.

Also, what velocity would I use? I worked out the velocity that the satellite should have if it was orbiting at that radius (it was about 7000 m/s, using Kepler's law of periods), but wouldn't that be tangential to the Earth?

Furthermore, how would I consider both the G.P.E. and K.E. ? I really don't get this part.

∆KE is 1/2 mv₂² - 1/2 mv₁², and PE you already know.

v₂ you can calculate from Newton's law of gravitation, the value of g at the Earth's surface, and the formula for centripetal acceleration

v₂ you can calculate from the radius of the Earth and the length of a day (assuming that the launch is, as Filip Larsen says, from the equator).

 

[minifhncc]

Hi again,

I understand what K.E. is, but I'm not sure how I'd use the value I get from the change in G.P.E. and the value I get from K.E.

Would I add them together?

v₂ you can calculate from Newton's law of gravitation, the value of g at the Earth's surface, and the formula for centripetal acceleration

Wouldn't you mean v₁?

Thanks

 

[tiny-tim]

I understand what K.E. is, but I'm not sure how I'd use the value I get from the change in G.P.E. and the value I get from K.E.

Would I add them together?

yes … that's the great thing about the different forms of energy, they're interchangeable, you just add them.

Wouldn't you mean v₁?

oops!

 

[minifhncc]

yes … that's the great thing about the different forms of energy, they're interchangeable, you just add them.

Okay thanks for that.

So, I'd find the orbital speed of the mass at the surface (v=(GM/r)^(1/2)) and then use this as v₁? But then how would I account for the length of a day using this method?

Thanks

 

[tiny-tim]

So, I'd find the orbital speed of the mass at the surface (v=(GM/r)^(1/2)) and then use this as v₁?

No! the rocket isn't in orbit on the surface, it's just going with the flow, like you and me.

 

[minifhncc]

No! the rocket isn't in orbit on the surface, it's just going with the flow, like you and me.

Umm okay, then does that mean v₁ = 0?

Sorry for all these questions that may appear elementary to you, I'm usually okay at Physics, but this one is really doing my head in :(

 

[tiny-tim]

Umm okay, then does that mean v₁ = 0?

No, if it's on the equator, it'll be going round the circumference of the Earth every 24 hours.

 

[minifhncc]

No, if it's on the equator, it'll be going round the circumference of the Earth every 24 hours.

Sorry I'm really not getting this then... :( If I'm not supposed to use v=sqrt(GM/r) as the orbit speed of the rocket at the surface of the Earth, then what value of v₁ am I supposed to use?

How would I take into account the daily orbit of the Earth's rotation (which I'd guess that you'd need to somehow need to use Period, T = 24 * 60 * 60 seconds).

Again I apologise if this seem really elementary to you, it's really starting to annoy me :(

EDIT: I've realized that you probably meant, to find v₁ that I use v^2/r = 9.8 (where r is the avg. radius of the Earth)... but the question still remains as to how I should take into account the Earth's daily motion...

 

[Filip Larsen]

As seen from outside the Earth (from an inertial reference frame not rotating with the Earth - like you watching a rotating desktop Earth globe) the Earth is rotating once per sidereal day (almost 24h) and anything standing still on a point on the equator will make a uniform circular motion. Knowing the rotation rate and circumference (or radius) of this motion you can calculate the tangential speed of the satellite before launch.

 

[tiny-tim]

For the PE you use the difference in GM/r (you've already done that).

For the KE you use the difference in 1/2 mv² …

v₂ is the final speed, which happens to be the orbital speed.

v₁ is the initial speed, which is when the satellite is in your garage.

Your garage goes round in a large circle every 24 hours … what speed is that?

 

[minifhncc]

Your garage goes round in a large circle every 24 hours … what speed is that?

Is it just v=(2*Pi*r)/T where r=avg radius of the Earth and T=24*60*60?

 

[tiny-tim]

Yup!

(though you may as well take the actual equatorial radius, if you're going to take it to the equator and launch it there )

 

[minifhncc]

Hey again,

I've got another question (which hopefully will be the last). For the given question, I got that the total energy required = 6.31*10^9 J could someone please confirm this answer?

Also, for the speed of the satellite on the Earth, when I use mv^2/r = 9.8, I get v=573 m/s (m=190kg, r = 6.37 * 10^6) and when I use v=(2*Pi*r)/T I get 463 m/s (approx) (r=6.37*10^6, T=24*60*60). Which way is correct?

I understand that it may be some errors dues to the rounding in the constants given... Could someone please clarify?

Thanks again :D

 

[Filip Larsen]

For the given question, I got that the total energy required = 6.31*10^9 J could someone please confirm this answer?

I get a result very close to that value.

Also, for the speed of the satellite on the Earth, when I use mv^2/r = 9.8, I get v=573 m/s (m=190kg, r = 6.37 * 10^6) and when I use v=(2*Pi*r)/T I get 463 m/s (approx) (r=6.37*10^6, T=24*60*60). Which way is correct?

The last one (the actual value is around 465 m/s because the rotation period of the Earth relative to the stars is around 4 minutes less than 24 h).

The first relation you mention does not make sense to me. I can't really guess how you derived it, but if you look at the physical dimension of the involved variables you can quickly conclude that is does not represent a physical valid relationship (unless the 9.8 has dimension energy per length ).

 

[minifhncc]

I get a result very close to that value.

That's reassuring :D

The first relation you mention does not make sense to me. I can't really guess how you derived it, but if you look at the physical dimension of the involved variables you can quickly conclude that is does not represent a physical valid relationship (unless the 9.8 has dimension energy per length ).

Ahh sorry. I messed up between centripetal acceleration (which is 9.8 m/s^2 at the surface), and force... In any case I don't think that applies here since the satellite at the surface is not in orbit.

All good now. Thanks a lot :D

 

[tiny-tim]

… I messed up between centripetal acceleration (which is 9.8 m/s^2 at the surface), and force... In any case I don't think that applies here since the satellite at the surface is not in orbit.

That's right …

v²/r = 9.8 would give you the speed at which Newton would have had to throw his apple to get it into orbit!