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.

 

[PJC01]

Firstly, in response to the first question about the properties of planets X and Y, the correct answer is B) Average density. This can be deduced from the equation for period of rotation which involves the mass and radius of the planet. Since the periods of rotation are the same, the ratio of mass to radius cubed must also be the same, which is equivalent to the average density being the same. This can also be seen by considering a smaller planet with a lower mass but higher density and a larger planet with a higher mass but lower density. Both would have the same period of rotation if their average densities are the same.

In regards to the second question about the collision of A and B in the spacecraft, your initial thoughts are correct. The velocities of A and B must be different in order for them to perform uniform circular motion. When A throws the box to B, the momentum of the system (A with the box and B) is conserved, but the individual momenta of A and B change. This means that A will have a different velocity after throwing the box and will no longer be able to perform uniform circular motion. The same goes for B. However, this does not necessarily mean that A and B will collide. If the box is thrown with the correct velocity, it will continue to perform uniform circular motion around B, preventing a collision between A and B. The velocity of the spacecraft will not be affected as it is a closed system and the external forces (centripetal forces) do not change.