Find the height if an orbiting satellite given only velocity.

[lubo]

Homework Statement

An Earth satellite is to be put into orbit at a height of 2 Mm above the Earth's surface. Determine the velocity it must have. Make the following assumptions: (a) the orbit is circular; (b) the Earth is a sphere of radius 6•37 Mm; (c) gravity decreases as an inverse square law.


(MY main question is to get an idea of how to find the height above the Earth's surface of the satellite, when you are only given the Velocity of a satellite in orbit).

Homework Equations

g' (acceleration due to gravity at a distance R to the centre of the earth)=g*r earth^2/R orbit^2

a=V^2/r orbit

The Attempt at a Solution

g'=?
g=9.81
r earth=6.37Mm
R orbit=6.37 + 2(height)

g' = 9.81*6.37^2/(2+6.37^2)
g'=5.682m/s^2

a=v^2/R therefore v^2=8.37*10^6 * 5.682 Therefore V=6896 m/s

However if I try to work back using V only V=6896 m/s and find the height 2Mm, I can only use g=9.81. Therefore I cannot find g' to allow me to find Ro (re + h). Then I could take Ro from re and I would have the height.

i.e. a=v^2/R = 6896^2/r Earth as I don't have R orbit = 6896^2/6.37*10^6 = 7.465m/s^2

If I try g' =mg then I do not have a mass.

If I try a=rw^2 and find w then I can only use re and a=9.81. Therefore I cannot find the g'= 5.682m/s^2

If I substitute g' with rw^2 and get Ro^3=g*r^2/w^2 I am still only using r of Earth and w from r and not a time period or R orbit

I hope all this makes sense, Thanks for any help.

 

[rock.freak667]

The Attempt at a Solution

g'=?
g=9.81
r earth=6.37Mm
R orbit=6.37 + 2(height)

g' = 9.81*6.37^2/(2+6.37^2)
g'=5.682m/s^2

a=v^2/R therefore v^2=8.37*10^6 * 5.682 Therefore V=6896 m/s

However if I try to work back using V only V=6896 m/s and find the height 2Mm, I can only use g=9.81. Therefore I cannot find g' to allow me to find Ro (re + h). Then I could take Ro from re and I would have the height.

i.e. a=v^2/R = 6896^2/r Earth as I don't have R orbit = 6896^2/6.37*10^6 = 7.465m/s^2

So what you did is that you found the value for g at the distance R_orbit, then used g =v²/R_orbit to get v. Then used v to get back R_orbit ?

If you need to get the velocity, why are you getting the height above the Earth's surface for?

Though if you needed to get 'h' using v you would just use

GMm/(R+h)² = mv²/(R+h)

mass of satellite m cancels out, M and G are constants and so is R.

 

[vela]

However if I try to work back using V only V=6896 m/s and find the height 2Mm, I can only use g=9.81. Therefore I cannot find g' to allow me to find Ro (re + h). Then I could take Ro from re and I would have the height.

i.e. a=v^2/R = 6896^2/r Earth as I don't have R orbit = 6896^2/6.37*10^6 = 7.465m/s^2

You can't arbitrarily use R_earth just because you don't have R_orbit.

If I try g' =mg then I do not have a mass.

This doesn't make sense at all. You have an acceleration on the lefthand side and a force on the righthand side of the equation. They can't be equal.

If I try a=rw^2 and find w then I can only use re and a=9.81. Therefore I cannot find the g'= 5.682m/s^2

If I substitute g' with rw^2 and get Ro^3=g*r^2/w^2 I am still only using r of Earth and w from r and not a time period or R orbit

I hope all this makes sense, Thanks for any help.

The problem is you're trying to find a number for an intermediate result, g', that you don't have enough information to find. You don't need to calculate the number, though.

The way you solved the problem initially was saying the acceleration due to gravity the satellite experiences isg' = g\frac{R_\mathrm{Earth}^2}{R_\mathrm{orbit}^2}This acceleration is also the centripetal acceleration needed for the circular orbit, so you have
g' = \frac{v^2}{R_\mathrm{orbit}}
Now if you had combined these two equations, you would have gotten
g\frac{R_\mathrm{Earth}^2}{R_\mathrm{orbit}^2} = \frac{v^2}{R_\mathrm{orbit}}
If, as in this problem, you're given R_orbit, you can solve for the speed v. On the other hand, if you were given v instead, you could solve the equation for R_orbit. In either case, you don't need to calculate the value of g'.

Avoiding the stumbling block you encountered is one of the reasons why your instructor has likely encouraged you to work out problems symbolically and to plug numbers in only near the end. I really encourage you to get into this habit. It'll save you a lot of headaches down the road.