Question

A satellite revolves in a circular orbit around a planet of density 2700kg/m3. The radius of the orbit is twice the radius of the planet. The universal gravitational constant is 6.67×10−11 in SI units. Find the period of the satellite.

Answer 1

I understand that this must use Kepler's Third Law, but I can't figure out how to use the density to find the mass of the planet. I've been using p = m/v, where V = 4/3*pi*r^3, but I haven't been successful in getting the right answer.

Answer 2

\[T = 2 \pi (\frac{a^3}{GM})^{\frac{1}{2}}\]
Where a is the semi major axis of the orbit (It's circular, so it's just the distance from the center of the planet. M is the mass of the planet.

Can you take it from here?

Answer 3

Did you derive that from Kepler's Third law? I thought the equation was \[T^2=\frac{ 4\pi^2a^3 }{ GM }\]

Answer 4

Oh wait. I see now haha.That's what I've been doing, however, the answer of 7234.2 is wrong.

Answer 5

That's not what I get when I evaluate it.

\[T = 2\pi (\frac{(2r)^3}{\frac{4}{3} \pi r^3 \rho})^{\frac{1}{2}} = (\frac{24 \pi}{G \rho})^{1/2}\]

Plugging in numbers, I get 20461.41944

Answer 6

I wasn't using 2r--why do you have to use that? But that number is the right answer so thank you!!

Answer 7

The 2r comes from the radius of the orbit being twice that the radius of the planet. So, a = 2r, if r is the radius of the planet.

Answer 8

Okay, I understand that now. Thanks again.
Would you be able to help me with another question I have?

Answer 9

Sure. What's up?

Answer 10

Okay here it is: A spherical, non-rotating planet with negligible atmosphere has mass m1 and radius r1. A projectile of mass m2≪m1 is fired from the surface of the planet at a point A with a speed vA at an angle α=30∘ with respect to the radial direction. In its subsequent trajectory the projectile reaches a maximum altitude at point B. The distance from the center of the planet to the point B is r2=(5/2)r1. Find the initial speed vA

Answer 11

So, we can easily solve for the radial velocity (the velocity directly away from the center of the planet).
\[\frac{1}{2}m\dot r ^2 = (\frac{5}{2} - 1)rmg = \frac{3}{2}rmg\]
where r dot is the radial velocity (time derivative on r). This just comes from conservation of energy:
\[\frac{1}{2}mv^2 = mgh\]
Just solve for r dot, and you should be able to do the rest easily.

Answer 12

So the answer should just be sqrt(3rg)?

Answer 13

That's the radial velocity, which is only a portion of Va. You'll have to use a trig function to find Va.

Answer 14

cos(30)= Va/Vradial, so Va = cos(30)*sqrt(3rg). Correct?

Answer 15

Your triangle is backwards, but close.

Va cos(30) = sqrt(3rg)

Answer 16

It says it's incorrect :/

Answer 17

csc(30)*sqrt(3rg)

or sqrt(3rg)/cos(30)

Says that's incorrect? Hmmm.

Answer 18

err, sec(30), not csc(30)

Answer 19

Yup. I put sqrt(3*r*g)/cos(30) and it was marked wrong...

Answer 20

Hmm, let me think about this then. I must be missing something.

Answer 21

Oh, did you mark m as m2?

Answer 22

m2 is the mass of the object. m1 is the mass of the planet. Since m2<<m1, the mass of the planet doesn't factor into the problem, to first order.

Answer 23

Or, I could read my own answer, and see that the mass isn't even in it. Haha

Answer 24

^That's what I was going to say haha.

Answer 25

The program this question is on denotes possible symbols to enter in the answer, and while not all are always used it's a good indication of what the answer should look like. For this problem, it includes big G, r1, and m1. We put little g in the answer...should it include big G somewhere instead or in addition to g?

Answer 26

Ah, the force of gravity is going to change in an appreciable way during the trajectory. At the top, it'll be 2/3 of the value at the surface. So I can't just cheap out on that part.

Answer 27

Try: sqrt(6*G*m1/5r)/cos(30)

If it's right, I'll explain how I got it.

Answer 28

\[\frac{ \sqrt{6*G*\frac{ m_1 }{ 5*r_1}} }{ \cos(30) }\]

Answer 29

Like that? Just want to make sure I'm inputting it in correctly.

Answer 30

yeah

Answer 31

As I wrote it there, it's incorrect haha.

Answer 32

I don't know if you need to expand G or not (The gravitational constant)

Answer 33

Hmmm...what am I missing. I used work energy to get that.

Answer 34

Very frustrating question.

Answer 35

Oh, I missed one of your messages earlier. You could try:
\[\frac{\sqrt{\frac{3Gm_1}{r}}}{Cos(30)}\]

I'm not sure how many attempts you have.

Answer 36

3 total, so only one more before the final check. (A friend and I are sharing accounts to get more chances, lol.) I'll try that now!

Answer 37

That was wrong too. Ugh.

Answer 38

Hmmm, strange.

Here's what we've tried so far, but I seem to be missing something:

\[\frac{1}{2}m\dot r^2 = mgh \rightarrow \dot r = \sqrt{3rg} = \sqrt{\frac{3Gm_1}{r}}\]

\[\int\limits_{r_1}^{\frac{5}{2}r_1}G\frac{m_1 m_2}{r^2}dr = 3G\frac{m_1 m_2}{5r_1} = \frac{1}{2}m_2 \dot r^2 \rightarrow \dot r = \sqrt{\frac{6Gm_1}{5r_1}}\]

Both with
\[V_a = \frac{\dot r}{Cos(30)}\]

Answer 39

Yup...and both those approaches make sense to me, but I have no idea why neither is the right answer.

Answer 40

Well, here's one more:
\[\dot r = \sqrt{\frac{4 G m_1}{3r_1}}\]

Which comes from:
\[\Delta V = G\frac{m_1 m_2}{\Delta r}\]

Remember to divide by cos(30)

Answer 41

but I'm not a fan of that approach. Depends on what they said in your class, I guess.

Answer 42

I personally wouldn't try it, as it's not correct, haha. But, I'm stumped, embarrassingly enough.

Answer 43

In truth, the website is so fickle that any one of your answers could be really close but off by a very small thing that it'll mark it as incorrect. I'm going to try that one anyway just for kicks.

Answer 44

Incorrect again..

Answer 45

I wonder if they wanted Cos(30) to be expanded.

\[Cos(30) = \frac{\pi}{6}\]

Answer 46

err
\[\frac{\sqrt{3}}{2}\]

Answer 47

Which would give answers of:
\[2 \sqrt{\frac{Gm_1}{r_1}}\]
and
\[2 \sqrt{\frac{2Gm_1}{5r_1}}\]

Answer 48

Both wrong. The trig is definitely right?

Answer 49

Yeah, cos(30) = sqrt(3)/2

Answer 50

From a friend, the answer is sqrt(5*G*m1/4*r1)...no clue how he got there though! Haha

Answer 51

.