# Space probe, planet, and sun .

1. Dec 5, 2003

### ConfusedStudent

Space probe, planet, and sun.....

Correcting test again, and I don't know where to start on this problem:

A space prob lies along a line between a planet and the sun so that the sun's gravitational pull on the probe balances the planets pull. The distance of the probe from the planet is 1.0 x 10^11m. The distance between the planet and the sun is 1.5 x10^12m. Find the mass of the planet. Mass of the sun = 2x10^30kg.

I got 0/15 on this problem so nothing I put down was right....help???

2. Dec 5, 2003

### gnome

what is the equation for the gravitational force between two objects?

Fg = ???

3. Dec 5, 2003

### ConfusedStudent

F1,2=[(G *m1*m2)/(r1,2^2)]*r1,2

That's what it says in my book, it looks weird when you type it.

But I tried this problem another way, and tell me what you think:

I used t^2=Cr^3 compared to the earth to get the period of the planet which I got was 6.53x10^15.

Then I used that in the eq:
T^2=(4*pie^2)/(G*Mp)*r^3
6.53x10^15m^2=(39.5)/(6.67x10^-11*Mp)*1.5x10^12m^3
and I got the mass to be 4.69x10^16 kg

Is this completely off??[?]

4. Dec 5, 2003

### gnome

Yes, it's completely off. The M in that equation (Kepler's Third Law) is MS, not MP. It's the mass of the sun. You can't use that equation to find the mass of the planet.

But, back to the other equation, the last r1,2 that you see there has a little ^ on top of it, right? That is a unit vector to indicate the direction of the force. It has no effect on the magnitude of the force, which is all we're concerned with here (because here we know that the two forces act along the same line, in opposite directions).

So, leave off that last r.

Now
1. call the planet X, so it's mass is MX, and the distance from the sun to the planet is rSX and the distance from the planet to the probe is rXp. And let's call the mass of the probe Mp.

2. You know that FSp = FXp

3. Find rSp

4. Can you set up an equation for the two forces using
F = GM1M2/(r122)?

5. Dec 5, 2003

### gnome

By the way, do you realize that r1,2 in the gravitational force equation is just the distance between the two objects?

(In other words, it is not the radius of the objects' orbits around the sun, which is what Kepler's equation is all about.)

6. Dec 5, 2003

### ConfusedStudent

I must really be dense, I've been looking at this screen for 15 min, I still don't get it.

7. Dec 5, 2003

### Staff: Mentor

The force of the sun on the probe equals the force of the planet on the probe. In other words (calling the Sun 1, the planet 2, and the probe 3):

F13 = F23

Now write what each force is using the gravity law:

FAB = G MAMB/(RAB)2

When you plug this into the first equation, G and M3 will cancel. You have the distances and the mass of the sun. The only variable left will be M2; solve for it.

Does this help at all?

8. Dec 6, 2003

### ConfusedStudent

Okay, this is what I got, and I hope it's right:

F(13)=F(23)

GM1M3/(R13^2)=GM2M3/(R23^2)

M1/(R13^2)=M2/(R23^2)

2x10^30/(1.4x10^12^2)=M2/(1.0x10^11^2)
M2= 1.02x10^28

Was it okay to assume the distance from the probe to the sun is the distance from the planet to the sun- the distance from planet to probe???

9. Dec 6, 2003

### Staff: Mentor

Yep.
Of course. I wouldn't call that an assumption; it's a given since they're all in a straight line.

10. Dec 6, 2003

### ConfusedStudent

Thank you so much...you're a life saver...