Why Do Planets with the Same Orbital Period Have the Same Average Density?

[abcd8989]

1) Two planets X and Y (sphere) both have 1 satelite revolving at low altitude about them. If the theri periods of rotation are found to be the same, which of the followings properties regarding X and Y is most likely to the the same?
A) Mass
B) Average density
C) Acceleration due to gravity at their surfaces
D) Radius
The answer is B. I don't know how this answer can be figured out.
My attempt before :
The centripetal force required for the satelite is provided by gravitation force
=> GMm / r ² = mv²/ r
=> v= sqrt( GM / r)
Period = 2 (pi) r / v
= 2 (pi) sqrt ( r³ / GM)
Their period are the same=> 2 (pi) sqrt ( r_x³ / GM)= 2 (pi) sqrt ( r_y³
=> r_x³ / M = r_y³ / M
Thus I guess the answer to be D. How should be answer be deduced correctly by appropriate concepts?


2) Suppose there is a large spacecraft revolving about the earth. There are 2 people, A and B, of same mass in the spacecraft each approaching other in opposite direction. A is carrying a box. The question is, how can collision of A and B be prevented by A's throwing the box to B?
First of all, prior to going into the question, I think their speed should be different. Since A(with a box) and B have different mass, they have to have differecnt velocity so that they can still perform uniform circular motion. Is it right? Moreover, for B to be able to receive the box, A should throw the box with a range of velocity such that the box can perform uniform circular motion around more or less the same radius as that of B, right?

As A throws the box to B, some momentum is transferred to from A to B via the box, right? However, in this case, the principle of conservation of linear momentum does not hold, since there is external forces acting on the system ( A with a box together with B), which is the centripetal forces (or gravitation forces). Hence, I wonder if the above consideration can still be valid. If yes, why? If no, what should be considered?

If A and B are not to collide, after throwing the box, A should have reverse velocity with v_A> V_B. However, after thinking in deep, I guess that after throwing the box, the velocity of A changes. And A can no longer perform uniform circular motion, and so is B, right? Eventually A and B would hit the spacecraft . Would the velocity of the spacecraft be affected?

Thanks a lot to answer me such conceptual troublesome questions!

 

[Yuqing]

You are essentially correct for 1 for your initial steps. The key here seems to be that both satellites are in low altitude orbit. We can equate the gravitational equations of the two planets as

$$\frac{M_{x}}{(r_{x} + d_{x})^{3}} = \frac{M_{y}}{(r_{y} + d_{y})^{3}}$$

where r is the radius of the respective planets and d is the altitude of the orbiting satellite. Because of the assumption that the satellites are in low altitude orbit, we can take $$d\approx0$$ which reduces the equality to

$$\frac{M_x}{(r_x)^{3}} = \frac{M_y}{(r_y)^{3}}$$

which is equivalent to

$$\frac{M_x}{\frac{4}{3}\pi(r_x)^{3}} = \frac{M_y}{\frac{4}{3}\pi(r_y)^{3}}$$
$$\rho_x = \rho_y$$

As for 2, the question is momentum conservation. I can only offer an intuitive description. The gravitational force shouldn't matter because the motion is perpendicular to the force and momentum in that direction is in fact conserved (This becomes more clear if you take the co-rotating frame of reference). I don't think the spacecraft is relevant for this problem and it merely provides a setting for the problem (which requires two people to be weightless). Heuristically, person A needs to throw the box with enough speed so that the faster of the two will reverse direction (or at least come to a stop).

 

[abcd8989]

You are essentially correct for 1 for your initial steps. The key here seems to be that both satellites are in low altitude orbit. We can equate the gravitational equations of the two planets as

$$\frac{M_{x}}{(r_{x} + d_{x})^{3}} = \frac{M_{y}}{(r_{y} + d_{y})^{3}}$$

where r is the radius of the respective planets and d is the altitude of the orbiting satellite. Because of the assumption that the satellites are in low altitude orbit, we can take $$d\approx0$$ which reduces the equality to

$$\frac{M_x}{(r_x)^{3}} = \frac{M_y}{(r_y)^{3}}$$

which is equivalent to

$$\frac{M_x}{\frac{4}{3}\pi(r_x)^{3}} = \frac{M_y}{\frac{4}{3}\pi(r_y)^{3}}$$
$$\rho_x = \rho_y$$
Thanks for your reply.However, I don't understand why the average density is the same while the radii are not the same. In fact, I discovered that my logic was wrong. $$\frac{M_x}{(r_x)^{3}} = \frac{M_y}{(r_y)^{3}}$$ does not necessarily imply their radii are the same, let alone multiplying 4/3 pi to them converting them to density.

 

[Yuqing]

It doesn't matter if the radii are different, it will be compensated by the fact that the mass will also be different. The equality is not between radii, it is between the mass divided by the cube of the radius which is essentially an equality between the density.