Calculating Energy for Satellite Orbit at 300km | Physics Homework Help

[charlie05]

Homework Statement

Satellite m=3 000kg, circular orbit at the height h = 300km above the Earth´s surface.

a/ calculate energy ( work ) for delivering satellites into orbit ?

b/ during 1 month the satellite approached the Earth of Δh=13,8 km. Prove that the radial component of velocity in comparison with the tangential component is insignificant.

c/ What the average resistance force is applied to the satellite during flight?

Homework Equations

I know: R_E = 6 370 km...radius of the Earth
M_E = 5,98 .10²⁴ kg...mass of the Earth
G = 6,67 . 10 ^-11 Nm²kg^-2...gravitational constant

The Attempt at a Solution

a/[/B]
W = E_p1 - E_p0
W = - (G*M_E* m) / 2( R_E + h) - (- (G*M_E* m) / R_E
W = G*M_E* m * (( R_E+2h)/ 2R_E( R_E +h))

with numbers: W = 9,81 . 10¹⁰ J

 

[gneill]

Okay. Did you consider the satellite's initial kinetic energy associated with the rotation of the Earth at the time of the satellite launch? Did you have a reason to ignore it?

You should have listed the equations you used to calculate the energies in the relevant equations section.

 

[charlie05]

yes...
W = ΔEp + ΔEk
ΔE_k = 1/2 mv² - 0

 

[gneill]

yes...
W = ΔEp + ΔEk
ΔE_k = 1/2 mv² - 0

So the Earth is not rotating?

 

[charlie05]

Yes, the Earth moves, but I think -change in kinetic energy can be ignored due to the small mass of the satellite in comparison to the Earth ?

 

[gneill]

Yes, the Earth moves, but I think -change in kinetic energy can be ignored due to the small mass of the satellite in comparison to the Earth ?

Probably something you should verify. What's the KE of the satellite while resting on the Earth's surface? Assume a launch from the equator.

 

[charlie05]

a circuit the Earth s = 40 000 km
time...24 hour
v = s/t = 40 000/24 = 1667 km/hour = 463 m/s

Ek = 0,5 * 3000 * 463² = 321,5 MJ

 

[gneill]

a circuit the Earth s = 40 000 km
time...24 hour
v = s/t = 40 000/24 = 1667 km/hour = 463 m/s

Ek = 0,5 * 3000 * 463² = 321,5 MJ

Okay. Does that represent a significant contribution to the work done? If not then you are justified in ignoring it.

 

[charlie05]

comparison ... 9,81.10⁷ MJ x 321,5 MJ...yes, I can ignore ΔE_k

 

[gneill]

comparison ... 9,81.10⁷ MJ x 321,5 MJ...yes, I can ignore ΔE_k

Good!

 

[charlie05]

super :-) so it is a/ , but I don't know how prove b/
radial velocity of the satellite is the centripetal force ?

 

[gneill]

super :-) so it is a/ , but I don't know how prove b/
radial velocity of the satellite is the centripetal force ?

Velocity is not force.

It's not clear to me what the import is of the Δh. The problem doesn't say whether this is due to the satellite's orbit becoming elliptical or if it remains circular but is shrinking in size over time. Looking at part (c) I'm leaning towards the latter, the orbital radius is shrinking over time due to some frictional force eating away at the KE of the orbit.

For (b) I suppose you could look at the definition of velocity since you have the radial displacement and the time period over which it occurred.

 

[charlie05]

maybe calculate the tangent to the path that the satellite executed for one month ... as a circuit track * time and compare it with the Δh?

path is circular...h = 300 km, R_E = 6 370 km...radius of the Earth, circumference = 2π( R+h)
in 1 month...s=t*circumference...but I don't know t :-( that way probably is not the way :-(

 

[charlie05]

v = √G*M/r

 

[gneill]

The satellite's R changed by Δh over one month. What average velocity does that give you?

What's the satellite's on-orbit velocity?

 

[charlie05]

v = √ 6,67.10^-11 * 5,98 . 10²⁴ / 6 370 000 + 300 000 = 7739 m/s

 

[charlie05]

vector v = vector radial component of velocity v_r + vector tangential component of velocity v_t

the numerical value...

 

[charlie05]

What should be the next steps, please?

 

[gneill]

What should be the next steps, please?

What precisely are you trying to determine?

 

[charlie05]

that radial component of the velocity is in proportion to the tangential insignificant...

during 1 month the satellite approached the Earth of Δh=13,8 km

can I compare flew a distance s = v*t with Δh?

 

[charlie05]

s _t = 7739 * 30 * 24 * 3600 = 20 * 10⁹ m = 20*10⁶ km

20.10⁶ km is disproportionately greater as 13,8 km ?

 

[gneill]

You want to compare velocities, not distances. You found a tangential velocity in post #16. What you need now is a value for radial velocity.

 

[charlie05]

Yes, it is true...

v_R = s/t = 13 800 / 30*24 = 19,6 km/hod ?

 

[gneill]

Yes, it is true...

v_R = s/t = 13 800 / 30*24 = 19,6 km/hod ?

Be careful with unit conversions. What you've calculated is meters per hour. You'll want to have both the tangential and radial velocities in the same units to make a comparison.

 

[charlie05]

yes :-(

v_r = 19,6 km/hod = 70,56 m/s

v_t = 7739 m/s

the tangential component of the velocity is about 100 times greater than the radial component...is it right?

 

[mfb]

What is "hod"? Hour, h?

13.8 km in a month is much slower than 70 m/s.

 

[gneill]

yes :-(

v_r = 19,6 km/hod = 70,56 m/s

v_t = 7739 m/s

the tangential component of the velocity is about 100 times greater than the radial component...is it right?

No. Your 19.6 km/hod (whatever a hod is...) is actually meters per hour. Check your conversion to meters per second. Write out the calculation.

Edit: Ha! @mfb got there before me!

 

[charlie05]

Yes, I'm sorry, I wanted to do it quickly :-(

v_r = 19,6 km/h = 5,44 m/s

v_t = 7739 m/s

the tangential component of the velocity is about 1400 times greater than the radial component.

 

[gneill]

No. Start again from scratch. First express Δh in meters, then express Δt in seconds. What do you get for each of those individually?

 

[mfb]

You can check your result with everyday comparisons: 5 m/s is a running speed. If you run for a month nonstop, clearly you run more than 14 kilometers?

 

[charlie05]

Δh = 13,8 km = 13,8 . 10³ m
I suppose ...1 month = 30 days = 30*24 h = 30*24*3600 s = 2 592 000 s

v_r = s/t = 13,8 . 103 / 2 592 000 = 0,005 m/s

 

[mfb]

That is right.

By the way, in English it is a decimal point, commas are used to group digits in large numbers. Example: 1,234,567.89 (about 1.2 millions)

 

[gneill]

Yes, much better. Only one significant figure, but since this is a ballpark approximation and comparison it will suffice. Now you can compare the tangential and radial velocities.

 

[charlie05]

Uffff :-) Thank you both for your patience - and of course help :-)

 

[charlie05]

comparison: v_t : v_r = 7739/0,005 = 1 547 800 / 1...the tangential component of the velocity is about 1 500 000 times greater than the radial component.

 

[gneill]

Good. So what is your conclusion for part (b)?

 

[mfb]

Right.

One comment on (a): While the kinetic energy on the ground is negligible, rockets don't really care about energies. They care about speed. The rotation of Earth is a significant point in rocket launches.

 

[charlie05]

yes, I undestand, thank you very much...