Question

Planet X of mass m1 orbits a Sun in uniform circular motion at a distance r1 and speed v1. The mass of the Sun is MS1 and its radius is RS1 as shown in the figure below.

Answer 1

2)
Consider another solar system in which Planet Y of mass m2 = 2 m1 orbits a Sun of mass MS2 = 4 MS1 and radius RS2 = 2RS1 at a distance r2 = 2r1. Further, suppose R2, the radius of Planet Y is twice R1, the radius of Planet X.

w2 < w1
w2 = w1
w2 > w1

I think its the last one because the 2's will cancel out but the M2 is 4 so it will be 2 after the canceling.

Also how would the speed compare to the original?

Answer 2

@Shadowys Again sorry for the constant badgering. I just want to make certain of my answers.

Answer 3

Nah, it's fine.
w being angular velocity?

Answer 4

I think so but it could also be weight.

Answer 5

The previous question asked for speed so it probably be angular velocity.

Answer 6

Yup, if it's angular velocity, then it would be the last one.
(I made the assumption that the radius of the planet is very small compared to the suns though)

Answer 7

As did I. Knowing that however, a follow up question asks that how would the velocity compare to the original?

v2 = (1/sqrt(2))v1
v2 = (1/2)v1
v2 = v1
v2 = sqrt(2)v1
v2 = 2v1

I think it is the last one because the angular velocity is bigger then w2>w1. All the other options would be less than w1 for example saying that w1=1

Answer 8

Wait. The first is the same. Sorry, I forgot to replace the m2 for the other side. It all cancels out to w1=w2

Answer 9

In thats case, v1=v2 then?

Answer 10

Since w1=w2, subbing \(w=\frac{v}{r}\), we get, \(v_2=2v_1\)

Answer 11

Oh I see. I made a silly assumption of thinking w=v.

I have just 1 last question. It involves the springs from an earlier post.

Suppose the top (bottom) spring is stretched (compressed) from its relaxed length by an amount δx = 0.082 m. If the spring constant of the top spring is three times that of the bottom spring and M = 0.25 kg and θ = 30o, what is k, the spring constant of the bottom spring?

k = 3.73 N/m
k = 4.98 N/m
k = 6.47 N/m
k = 7.47 N/m
k = 14.9 N/m

I really have no idea on how to start it. I have F = mg(cos30) but not sure on where togo from there.

Answer 12

Well, since it's frictionless, we can skip the y-axis and focus on the x-axis.
Since it's not moving, i.e. a=0,
taking downwards as positive, and along the incline,
\(\Sigma F_x =0\)
\(F_gcos 30- (F_{s(top)}+F_{s(bottom)}) =0\)
Well, from here sub everything you know in and voila! :) you get k

Answer 13

note: Sorry, it's
\(F_gsin 30−(F_{s(top)}+F_{s(bottom)})=0\)

Answer 14

How do I find Top and F bottom respectively?

Answer 15

well, from Hooke's law, \(F_s=kx\), since \(k_{top}\)=3\(k_{bottom}\) and their extensions will be same.

Answer 16

So is the extension .082?

Answer 17

Yup! for both springs.

Answer 18

I feel like I am doing this wrong