Gravitation and planetes: exercise

AI Thread Summary
An astronaut on two identical mass planets, A and B, conducts experiments by throwing a ball vertically, achieving different maximum heights of 25 cm and 4 m, respectively. The ratios of the planets' radii and densities are calculated, revealing that the radius ratio is 1:4 and the density ratio is 64:1, assuming uniform density. The discussion also addresses the effects of planetary rotation on gravitational acceleration, particularly at the equator versus the pole. Participants clarify the relationship between gravitational and centripetal acceleration, emphasizing the need for separate equations for each location on the planet. The conversation concludes with a focus on understanding the gravitational forces at play in the context of the astronaut's experiments.
RHK
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This is my first topic. I'm sorry for my english: that is not my language.
On the planet A, sferically symmetric and with no atmosphere, an astronaut on one pole throws in vertical upwards, a little ball giving to it a certain velocity: this ball reaches the maximum height of 25 cm.
The same thing is done on another planet, B, equal in mass to the planet A, and then the ball reaches the maximum height of 4 m.
Requests:
(i) to calculate the planetes radii ratio
(ii) to calculate the planetes densities ratio, with the hypothesis that the planets are uniform.

Successively, the astronaut make the same experiment on the B planet, this time on the equator, verifying that the ball reaches the height of 8 m.

(iii) calculate the rotation period of the planet, supposing that the mass of the planet is 10^27 g and its radius is 5000 km.

Temptative of solution:
(i) We know that E=mgh, so h=E/mg. Writing down the same equation for the two planetes, the following ratio is obtained: (hA/hB)=(R_A/R_B)^2 = 1/16

(ii) The density rho is writable as rho=M/V, where V=(4pi)/R^3, than (rhoA/rhoB)=(R_B/R_A)^3=64

Is this right? And then, how can i proceed for the third point?
Thanks a lot.
 
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the following ratio is obtained: (hA/hB)=(R_A/R_B)^2 = 1/16
......

chech the working in obtaining the above. I do not think it is correct if the densities of the two planets are not the same.
 
Thread moved to homework forums.
 
grzz said:
the following ratio is obtained: (hA/hB)=(R_A/R_B)^2 = 1/16
......

chech the working in obtaining the above. I do not think it is correct if the densities of the two planets are not the same.

What do you mean? How can density enter to play?
Thanks
 
RHK said:
What do you mean? How can density enter to play?
Thanks
If you show how you got the ratio of the heights then I can try to find if the method is correct.
 
This is my temptative for the first point:

E_A=m*g_A*h_A, so: h_A=E_A / m*g_A

E_B=m*g_B*h_A, so: h_B=E_B / m*g_B

Then it's possible to write: h_A / h_B = g_A / g_B

where g=(G M)/R^2 and so: (hA/hB)=(R_A/R_B)^2 = 1/16
What do you think about?
 
Note that M is the mass of the planet. but there are TWO planets and you have to distinguish between them - planet A and planet B.
 
I'm sorry: in the exercise text above the two planets masses are equal.
I've fixed the text.
 
Then I think that i and ii are correct.
As regards iii note that the rotation of the planet causes the acc due to gravity to reduce from that at the pole.
Use F = ma where a is the centripetal acc of a point on the equator towards centre of planet.
 
  • #10
Yes, i also thought at that.
But i can not start the problem: F=ma for the centripetal acceleration, where a=R*w^2 and w is related to the period. But i can not proceed...
 
  • #11
Is not the centripetal acc the reason that the acc due to gravity reduced by half from that at the pole?
 
  • #12
grzz said:
Is not the centripetal acc the reason that the acc due to gravity reduced by half from that at the pole?

Yes, of course it is.
But i don't get the point: I have to build another system with two equation, one for the pole and another one for the equator?
 
  • #13
Is it right to suppose a new system, with other two heights (a system for the same planet, with different heights at the pole and the equator)?
In this case i think that at the equator the acceleration is g+a_c isn't it?
 
  • #14
RHK said:
i think that at the equator the acceleration is g+a_c isn't it?

Since g_{equator} = (1/2)g_{pole}
then g_{pole}= 2R\omega^{2}
 
  • #15
I don't understand just a thing: the acceleration at the equator is ONLY the centripetal one? Why?
 
  • #16
g_{equatir} = GM/(R^{2}) - R\omega^{2}
 
  • #17
That's ok :)
Thanks a lot
 
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