What is the Mass of the Planet?

[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?

 

[gnome]

Start with:

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

F_g = ?

 

[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??[?]

 

[gnome]

Yes, it's completely off. The M in that equation (Kepler's Third Law) is M_S, not M_P. 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 r_1,2 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 M_X, and the distance from the sun to the planet is r_SX and the distance from the planet to the probe is r_Xp. And let's call the mass of the probe M_p.

2. You know that F_Sp = F_Xp

3. Find r_Sp

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

 

[gnome]

By the way, do you realize that r_1,2 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.)

 

[ConfusedStudent]

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

 

[Doc Al]

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:

F_AB = G M_AM_B/(R_AB)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?

 

[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?

 

[Doc Al]

Originally posted by ConfusedStudent
Okay, this is what I got, and I hope it's right:

Yep.

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?

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

 

[ConfusedStudent]

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