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BobbyJones

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The velocity v of an orbiting satellite is 2.862 X10^4 km/hr. calculate:

a)The height 'h' (in km) of the orbit above the earth's surface.

b) The period of rotation of the satellite giving the answer in hr-min-sec.


DATA: radius of earth is 6370km

Gravity is 9.81 m/s



Can someone point me in the right direction please.
 
BobbyJones said:
The velocity v of an orbiting satellite is 2.862 X10^4 km/hr. calculate:

a)The height 'h' (in km) of the orbit above the earth's surface.

b) The period of rotation of the satellite giving the answer in hr-min-sec.


DATA: radius of earth is 6370km

Gravity is 9.81 m/s



Can someone point me in the right direction please.
Click to expand...

Assuming You are looking at circular motion:

1) What is the centripetal acceleration of the satellite?

2) What is the gravitational acceleration of the satellite?

equate 1) and 2) to solve for 'h'.

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
 
The acceleration due to gravity 'g' at the distance r to the centre of the earth is given by


g' = (g x d^2)/ r^2

Where g = gravity at the earths surface (9.81 ms^-2) and 'd' is the mean radius of the earth (6370km)



Thats all the info I have, do you know which formulas I need to work out the centripetal force because I've found this one:

F = (mv^2)/r .....where I think r= 6370000, v = 2.862x10^4, but m = mass, but do I take it, it means the mass of the earth?

If so the formula I've found for that is m= (ar^2)/g = (9.81 x 6.4x10^6)/6.67x10^-11

Am I going off on a tangent here, or am I on track?
 
BobbyJones said:
The acceleration due to gravity 'g' at the distance r to the centre of the earth is given by


g' = (g x d^2)/ r^2

Where g = gravity at the earths surface (9.81 ms^-2) and 'd' is the mean radius of the earth (6370km)



Thats all the info I have, do you know which formulas I need to work out the centripetal force because I've found this one:

F = (mv^2)/r .....where I think r= 6370000, v = 2.862x10^4, but m = mass, but do I take it, it means the mass of the earth?

If so the formula I've found for that is m= (ar^2)/g = (9.81 x 6.4x10^6)/6.67x10^-11

Am I going off on a tangent here, or am I on track?
Click to expand...

use g' = v^2/r
 
We got told to use this formula g' = (g x d^2)/ r^2


But if g' = v^2/r = (2.862x10^4)^2/6370000 = 128.58


I've worked out b) to be 23 minutes 31 seconds, can someone confim if I'm right. I'm just struggerling with the height from earth.
 
BobbyJones said:
We got told to use this formula g' = (g x d^2)/ r^2


But if g' = v^2/r = (2.862x10^4)^2/6370000 = 128.58 Incorrect


I've worked out b) to be 23 minutes 31 seconds, can someone confim if I'm right. I'm just struggerling with the height from earth.
Click to expand...

r = 6370000 + h
 
BobbyJones said:
Therefore g' = (2.862x10^4)^2/6370000 + h


So how do I find 'h', I'm still in the dark.


Or is it

h = (((2.862x10^4)^2)/g') - 6370000 ? I'm still left with 2 variables!
Click to expand...

g' = g * [6370000/(6370000+h)]2 → g * [6370000/(6370000+h)]2 = v2/(6370000+h)

You know values of 'g' and 'v' - only unknown left is 'h'.
 
Last edited by a moderator:
Your saying

g' = g * [(6370000+h)/6370000]2 = v2/(6370000+h)

9.81 * [(6370000+h)/6370000]2 = (2.862x10^4)2/(6370000+h)

Would this then be

(6370000+h)* [(6370000+h)/6370000]2 = (2.862x10^4)2/9.81

How do I rearrange from here to get the 'h' on it's own?
 
BobbyJones said:
Your saying

g' = g * [(6370000+h)/6370000]2 = v2/(6370000+h)

9.81 * [(6370000+h)/6370000]2 = (2.862x10^4)2/(6370000+h)

Would this then be

(6370000+h)* [(6370000+h)/6370000]2 = (2.862x10^4)2/9.81

How do I rearrange from here to get the 'h' on it's own?
Click to expand...

There was a mistake above - it should as follows:

g' = g * [6370000/(6370000+h)]2 = v2/(6370000+h)
 
The velocity v of an orbiting satellite is 2.862 X10^4 km/hr. calculate:

a)The height 'h' (in km) of the orbit above the earth's surface.

b) The period of rotation of the satellite giving the answer in hr-min-sec.


DATA: radius of earth is 6370km

Gravity is 9.81 m/s



Can someone point me in the right direction please.

The satellite velocity in a circular orbit is given by Vc = sqrt(µ/r) where Vc = the velocity in m/sec, µ = the earth's gravitational constant = gR^2 = (3.980x10^14)m^3/sec^2 and r =the radius of the orbit in meters. R = 6,370,000m.

Solving for r and subtracting 6,370,000m yields the orbiting altitude.

Applying your numbers into Vc = sqrt[gR^2/r] yields an orbital radius of 6298.7km, less than the earth's radius.


The orbital period of a satellite in seconds derives from T = 2(Pi)sqrt(r^3/µ).

With the hypothetical orbital radius of 6298.7km, the orbital period becomes 4977.6sec = 1hr-22min-57.6sec.

Using the given earth radius of 6370km and a minimum orbital altitude of 100km(~62 miles), the required orbital velocity becomes 7844m/sec = 2.8234x10^4km/hr.

The orbital period at 100km altitude is then 5183 sec. = 1hr-26min-23sec.

Clearly, your satellite orbital velocity of 2.862x10^4 is incorrect, for earth at least.
 
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That is all the information I have been given. I've been looking at all different formulas, but I always end up with two unknown variables. Is it possible to work out the circumference from the velocity then I can work out the radius that way and minus the radius of the earth?
 
BobbyJones said:
That is all the information I have been given. I've been looking at all different formulas, but I always end up with two unknown variables. Is it possible to work out the circumference from the velocity then I can work out the radius that way and minus the radius of the earth?
Click to expand...

Wheredid you get this problem? As Tchrwill indicated that the orbital radius of the satellite is less than the mean radius of earth!!

So the satellite is travelling too slow for it to stay up!
 
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