Satellite Orbital Speed at 10% Gravity

[TonkaQD4]

(a) At what height above the Earth is the acceleration due to gravity 10% of its value at the surface?
(b) What is the speed of a satellite orbiting at that height?

Note: _e means earth

g = { g_e / [1+(h/R_e)^2] }

At height 0, g = 9.83m/s^2
R = 6400km

I know that we need to isolate "h". I don't quite understand how to start this problem though. Where do we plug in that 10% or what exactly does that 10% mean? I guess the wording of the question is what makes this problem confusing. Help me start this please.

 

[Bill Foster]

The acceleration of a mass at the surface is $$a_0$$

The acceleration of a mass at some height x above the surface is $$a_1$$

You want to know what x is when $$\frac{a_1}{a_0}$$=0.1

The rest is simple algebra.

 

[TonkaQD4]

That doesn't help me out.

Any other suggestions out there.

 

[dotman]

Maybe think about it this way... you have a formula that gives you some g for any value of h that you want to plug in.

Unfortunately, you need a formula that gives you the h for some value of g. That's ok, you can plug in your value for g (once you figure it out) and manipulate the equation.

Maybe if you saw Bill's equation with g's instead of a's:

g1 / g0 = 0.1 , where g0 is the acceleration due to gravity at h = 0. You can solve this for g1, which will be, in this case, 10% (0.1) of the acceleration due to gravity at h = 0.

Hope this helps.

 

[TonkaQD4]

Should I use the Universal Law of Gravitation instead?

F = GmM/r^2

Still not sure what to do then? I feel like I am making this problem harder than it really is.

 

[dotman]

Well, you probably are making this problem a bit harder than it is, but everyone does that until it 'clicks'.

Have you drawn a picture? I get the impression that you're trying to plug numbers into formulas, but you're not sure which formulas to use or why because you don't have a physical picture of what's going on. I would suggest drawing a picture. What happens to the gravitational force as you move away from the surface of the earth?

Now, as for whether or not you should use the formula for the gravitational force, you already are using it-- you just don't know it yet. Check out the following derivation, which I believe is accurate (I'm slightly out of practice :-)-- if anyone notices anything, please jump in. Now,

$$F = ma = \frac{GM_{e}m}{r^2}$$

and since we're calling the acceleration due to gravity g, not a,

$$F = mg = \frac{GM_{e}m}{r^2} \Rightarrow{g = \frac{GM_{e}}{r^2}}$$

So now, at the surface of the earth, $r = R_e$, so

$$g_e = \frac{GM_{e}}{{R_e}^2}$$

Now, at some height above the surface of the Earth $R_e$, $r = R_e + h$, so the acceleration due to gravity there would be

$$g = \frac{GM_{e}}{r^2} = \frac{GM_{e}}{(R_e+h)^2}$$

Now, using these two results, we can calculate the ratio $\frac{g_e}{g}$:

$$\frac{g_e}{g} = \frac{\frac{GM_{e}}{{R_e}^2}}{\frac{GM_{e}}{(R_e+h)^2}} = \frac{(R_e + h)^2}{{R_e}^2} = \frac{{R_e}^2(1+\frac{h}{R_e})^2}{{R_e}^2} = (1+\frac{h}{R_e})^2$$

so

$$\frac{g_e}{g} = (1+\frac{h}{R_e})^2 \Rightarrow g = \frac{g_e}{(1+\frac{h}{R_e})^2}$$

Which is the formula (almost; I think you may have mis-typed it) you were originally going to use. And you should, it will help you with part (a), although I have shown an intermediate result that may actually be slightly more useful.

The math is helpful, but try to picture what's going on here, to help you decide which equations you should use. Draw a picture, if you haven't. All you're trying to do is find out how high you need to be to experience 10% of the gravitational force as you would on the surface.

Hope this helps.

 

[Saladsamurai]

The acceleration of a mass at the surface is $$a_0$$

The acceleration of a mass at some height x above the surface is $$a_1$$

You want to know what x is when $$\frac{a_1}{a_0}$$=0.1

The rest is simple algebra.

That doesn't help me out.

Any other suggestions out there.

This certainly does help you out. $$\frac{a_1}{a_0}=.10$$

but a_0= acceleration due to gravity or just $$g_{on earth}$$

...so what is g in terms of G,M,m and R?

Hint: $$F_g=\frac{GMm}{r^2}$$ ... but what else is F_g equal to? Hint 2: Its a force!
Casey

 

[TonkaQD4]

Thanks. Figured it out.

 

[Saladsamurai]

The answer is 13838577.03 meters. hahahaha

This thread is over a year old; the joke would appear to be on you