3 Physics Questions

**kenan** · Nov 18th 2007, 10:34 AM

1. You want to launch a satellite into a circular orbit at an altitude of 16,000 km (above Earth's surface). What launch speed will be required?

A: 1.0 x 10^4 m/s

2. In a joint international effort, two rockets are launched from Earth's surface. One has an initial velocity of 13 km/s and the other 19 km/s. How fast is each moving when it crosses the Moon's orbit (3.84 x 10^8 m)?

6.8 km/s; 15 km/s

3. A 460 kg satellite is launched into a circular orbit and attains an orbital altitude of 850 km above Earth's surface. Calculate the additional energy and speed required for the satellite to escape.

2.2 x 10^9 J; 3.1 km/s

**topsquark** · Nov 18th 2007, 02:22 PM

Originally Posted by kenan

1. You want to launch a satellite into a circular orbit at an altitude of 16,000 km (above Earth's surface). What launch speed will be required?

A: 1.0 x 10^4 m/s

I tend to forget formulas. There is probably a nice, simple equation you can use to solve this, but here is a derivation:

We will need to account for the speed necessary for the satellite to go into a circular orbit, so let's find that speed first.

We know that the only force to act on the satellite is the Earth's gravity, so the force of gravity is equal to the centripetal force on the satellite:
$F_G = F_c$

$\frac{GmM_E}{r^2} = \frac{mv^2}{r}$
where m is the mass of the satellite and r is the distance from the center of the Earth to the satellite. (r is NOT 16000 km!)

Solving for v I get:
$v = \sqrt{\frac{GM_E}{r}}$
(You can calculate out what number this is, I'm going to leave it as a formula for now. As a reference, it's about $4200~m/s$ using $R_E = 6.367\times 10^6~m$ .)

Now, for the launch. We know what speed we want at the end, thus we know what final kinetic energy we want left over.

So I'm going to set an origin where the rocket takes off and let +y be upward. Gravity is a conservative force, so there are no non-conservative forces present. (I'm assuming the satellite is just thrust upward all at once instead of being in a rocket thrusting it upward over time.)

So:
$W_{nc} = \Delta E$

$\Delta KE + \Delta PE = 0$

$\frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 + -\frac{GmM_E}{r} - -\frac{GmM_E}{r_0} = 0$
(where I have set the 0 point for gravitational potential energy to be at infinity.)

We know the final speed, v, and we want $v_0$ , (and note that $r_0 = R_E$ , not 0.)
$\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 - \frac{GmM_E}{r} + \frac{GmM_E}{r_0}$

$v_0 = \sqrt{v^2 + 2GM_E \left ( \frac{1}{r_0} - \frac{1}{r} \right )}$

You can now plug your numbers in and get the answer. I'm going to stick that ugly expression in for v and see what comes out:
$v_0 = \sqrt{ \left ( \sqrt{\frac{GM_E}{r}} \right )^2 + 2GM_E \left ( \frac{1}{r_0} - \frac{1}{r} \right )}$

$v_0 = \sqrt{ \frac{GM_E}{r} + 2GM_E \left ( \frac{1}{r_0} - \frac{1}{r} \right )}$

$v_0 = \sqrt{GM_E \left ( \frac{2}{r_0} + \frac{1}{r} \right )}$

So I get:
$v_0 = 11963~m/s$
which is more or less what your given answer is.

-Dan

**topsquark** · Nov 18th 2007, 02:32 PM

Originally Posted by kenan

2. In a joint international effort, two rockets are launched from Earth's surface. One has an initial velocity of 13 km/s and the other 19 km/s. How fast is each moving when it crosses the Moon's orbit (3.84 x 10^8 m)?

6.8 km/s; 15 km/s

This is pretty much the same setup as the other problem, but this time we are solving for v, not $v_0$ . So I'm just going to pluck out one of the equations from my previous post.
$\frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 + -\frac{GmM_E}{r} - -\frac{GmM_E}{r_0} = 0$
(This came from the work-energy equation.)

We are looking for v so
$v = \sqrt{v_0^2 + \frac{2GM_E}{r} - \frac{2GM_E}{r_0}}$

$v = \sqrt{v_0^2 + 2GM_E \left ( \frac{1}{r} - \frac{1}{r_0} \right )}$

You can verify that this equation gives the correct answers.

-Dan

**kenan** · Nov 18th 2007, 02:50 PM

thanks a lot but for some reason im not getting the right answer in the first question, im getting 19971629.88. For $r_o$ i use 6.38 x 10^6, for r im using 2.238 x 10^7, G is 6.67 x 10^-11 and $M_E$ im using 5.98 x 10^24

Edit: well it seemed to work when i used the equation before your final one

**topsquark** · Nov 18th 2007, 03:25 PM

Originally Posted by kenan

3. A 460 kg satellite is launched into a circular orbit and attains an orbital altitude of 850 km above Earth's surface. Calculate the additional energy and speed required for the satellite to escape.

2.2 x 10^9 J; 3.1 km/s

I don't agree with your book here.

First let's find the speed of the orbiting satellite.
$F_G = F_c$

blah blah blah.

I already did this in the first problem, so just jump to the end:
$v = \sqrt{\frac{GM_E}{r}}$ <-- Again, you can calculate a number for this if you want.

Now we need to know how much more speed is required to make this thing escape the Earth altogether. What this means is, we want to give the satellite an additional amount of energy such that the satellite goes all the way out to infinity. Again we turn to the work - energy theorem:
$\frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 + -\frac{GmM_E}{r} - -\frac{GmM_E}{r_0} = T$
(T is the kinetic energy we are adding to the satellite to make it go meet the Enterprise. Since we have this energy from an outside source, this is considered non-conservative work.)

Our v from the previous equation is now the $v_0$ for this equation. Additionally r from the last equation becomes $r_0$ for this one. And we are going to put r as infinity (in practical terms, a very very very verrry large number). So the r term vanishes (becomes effectively 0.) Finally, we want just enough energy to do this. So we want v = 0 m/s as well.

$T = - \frac{1}{2}mv_0^2 + \frac{GmM_E}{r_0}$

Again, you can calculate a number for this, but I'm going to stick that expession in again:
$T = - \frac{1}{2}m \left ( \sqrt{\frac{GM_E}{r_0}} \right )^2 + \frac{GmM_E}{r_0}$

$T = - \frac{1}{2}m \cdot \frac{GM_E}{r_0} + \frac{GmM_E}{r_0}$

$T = GmM_E \left ( \frac{1}{r_0} - \frac{1}{2r_0} \right )$

$T = \frac{GmM_E}{2r_0}$

I get $T = 1.27108 \times 10^{10}~J$ . I've checked and rechecked my checking, and I can't find an error in my work.

Now, all this excess energy goes into kinetic energy, so
the total kinetic energy of the satellite just after the energy has been added is
$KE_{tot} = KE_0 + T = \frac{1}{2}mv_0^2 + \frac{GmM_E}{2r_0}$

$KE_{tot} = \frac{1}{2}m \left ( \frac{GM_E}{r_0} \right ) + \frac{GmM_E}{2r_0}$

$KE_{tot} = \frac{GmM_E}{r_0}$

And since this is a kinetic energy we can find out how fast the satellite is moving after the extra energy is added:
$\frac{1}{2}mv^2 = \frac{GmM_E}{r_0}$

$v = \sqrt{\frac{2GM_E}{r_0}}$

And we want the extra speed required, so
$\Delta v = \sqrt{\frac{2GM_E}{r_0}} - \sqrt{\frac{GM_E}{r_0}}$

$\Delta v = (\sqrt{2} - 1) \sqrt{\frac{GM_E}{r_0}}$

I get $\Delta v = 3.30052~km/s$

-Dan

**topsquark** · Nov 18th 2007, 03:32 PM

Originally Posted by kenan

thanks a lot but for some reason im not getting the right answer in the first question, im getting 19971629.88. For $r_o$ i use 6.38 x 10^6, for r im using 2.238 x 10^7, G is 6.67 x 10^-11 and $M_E$ im using 5.98 x 10^24

Edit: well it seemed to work when i used the equation before your final one

You know, I was just thinking it was odd that I had made exactly the same mistake at first, then I realized I had written the equation wrong. (You're supposed to catch me when I do that!!

) I have fixed the error. I had been adding the two terms under the radical when I should have been multiplying them.

-Dan

**kenan** · Nov 18th 2007, 03:55 PM

You have helped me out so much today
Thanks a lot for all your time

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