Mysterious Planet: Approximating its Density

  • Thread starter Thread starter tandoorichicken
  • Start date Start date
  • Tags Tags
    Density Planet
AI Thread Summary
The discussion revolves around calculating the density of a mysterious planet with no atmosphere, which has a satellite orbiting it every 1.50 hours. Participants note that while the orbital period is provided, additional information like the planet's radius or mass is typically necessary for density calculations. A proposed solution involves equating gravitational force to centripetal force, leading to a formula for density based on the orbital period. The derived equation suggests that density can be expressed as a function of the period, specifically as ρ = (1/G)(2π/T)². This approach offers a method to estimate the planet's density despite the lack of direct measurements.
tandoorichicken
Messages
245
Reaction score
0
A strange new planet that has no atmosphere has a satellite that orbits very close to the planet's surface with a period of 1.50 hours. What is the approximate density of the planet? (Assume that the radius of orbit equals the radius of the planet.)
 
Physics news on Phys.org
Maybe I'm wrong...but it sounds like something is missing here:
It only gives you the time it takes to complete a revolution?
I would think u either need to know the planet's radius, mass, or ship velocity to calculate the density...

Because, say a person drives a car around the Earth's equator, and someone else flies a plane around the same distance. Without knowing velocity of the objects, the period of each revolution would be different, even though its the same planet, hence same density...
 
How about just equating the gravitational force to the centripetal force?
\frac{mGM}{R^2} = m\omega^2R

and since \rho = \frac{M}{R^3} and v = Rω and T = \frac{2\pi}{\omega} you can solve for density as a function of period. So I think:

\rho = \frac{1}{G}\left(\frac{2\pi}{T}\right)^2
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Calculating Angle & Speed to Reach Planet's Moon from Station Orbit
Replies
1
Views
2K
Calculate magnetic dipole of other planets relative to the Earth
Replies
4
Views
1K
Find Density Given Period of Orbit
Replies
12
Views
5K
From circular orbit to elliptical orbit
Replies
3
Views
2K
Calculating Motion of a Test Balloon on a Methane Planet
2
Replies
62
Views
4K
Calculating Density of a Planet Based on Close Orbit Satellite Period
Replies
1
Views
2K
Calculation of Speed for Message Packet on Small Planet
Replies
4
Views
2K
Doubt about the elliptic orbits of the planets
Replies
1
Views
1K
Planets and satellites, law of periods circular orbit.
Replies
15
Views
2K
Two planets of equal density, what do they share?
Replies
7
Views
5K

Hot Threads

Recent Insights

Back
Top