Is the Time of Orbit for a Satellite Above Mars Really 800 Million Years?

  • Thread starter Thread starter Xanthippus
  • Start date Start date
  • Tags Tags
    Orbit Time
AI Thread Summary
The discussion centers on calculating the orbital period of a satellite around Mars, using the equation R^3/T^2 = 1x10^13 m^3 s^(-2). Initially, an incorrect calculation suggested an orbital time of 800 million years, which raised confusion about the acceleration involved. After reevaluating the math, the correct orbital period was determined to be approximately 2529.8 seconds, indicating a much faster orbit. Participants acknowledged the error in the initial calculations and clarified the correct approach. The conversation highlights the importance of accurate mathematical application in orbital mechanics.
Xanthippus
Messages
2
Reaction score
0

Homework Statement


We have been asked to work out the approximate time of orbit, T, of a satellite orbiting above the centre of Mars, radius 4000km.

Homework Equations


We have been given the equation:

R^3/T^2 =1x〖10〗^13 m^3 s^(-2)

The Attempt at a Solution


When using the equation:

T=√((4x〖10〗^6 )^3/〖1x10〗^13 )

I get an answer of 2.53 x 10^16 seconds and I don't understand how something that seems to be accelerating so fast turns out to be taking 800 million years to orbit mars. Is my math just totally wrong?

EDIT: That maths was wrong I'm pretty sure, I now have an answer of 2529.8 seconds which is pretty fast for an orbit of Mars but 1x10^13 m^3 s^-2 is a hell of an acceleration.
 
Last edited:
Physics news on Phys.org
Hi Xanthippus! :smile:
Xanthippus said:
EDIT: That maths was wrong I'm pretty sure, I now have an answer of 2529.8 seconds …

I think you're √10 out :wink:
 
I don't see where...

T=√((〖4x10〗^6 )^3/〖1x10〗^13 )

T=√(〖64x10〗^18/〖1x10〗^13 )

T=√(〖64x10〗^5 )

T=2529.8s
 
oops!

sorry, you're right! :redface:
 

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

Change in momentum for a satellite in circular orbit
Replies
6
Views
3K
Help with circular velocity question (unit conversions?)
Replies
2
Views
2K
Time that a comet spends inside Earth's orbit
Replies
2
Views
2K
Find Density Given Period of Orbit
Replies
12
Views
5K
Satellite Motion Homework: Find 2nd Satellite Speed
Replies
2
Views
2K
Finding r of a Jovian-synchronous orbit and escape velocity
Replies
7
Views
2K
A 1 500kg satellite is in orbit 250km above earth's surface.
Replies
14
Views
3K
What Altitude Should a Geosynchronous Satellite Orbit Mars?
Replies
2
Views
2K
Solving Satellite Orbit Problem: Finding Time in Hours
Replies
3
Views
2K
What Are the Formulas for Calculating Height in Synchronous Orbit?
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top