Bypass time of satelite arround Mars

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Homework Help Overview

The discussion revolves around calculating the bypass time of a satellite orbiting Mars at an altitude of 150 km, with the radius of Mars given as 3400 km. Participants are exploring the necessary parameters and equations involved in orbital mechanics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equation for orbital period and question the values for gravitational acceleration and the units used for radius and height. There is an exploration of the need for Mars-specific gravitational values and the implications of using different units.

Discussion Status

There is active engagement with participants providing feedback on unit consistency and formula correctness. Some participants have identified potential errors in the original calculations and are suggesting corrections, while others are still uncertain about the appropriate formula to use.

Contextual Notes

Participants are navigating the complexities of gravitational constants specific to Mars and the necessity of unit conversions between kilometers and meters. There is an acknowledgment of the need for accurate values to proceed with the calculations.

antoman
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Homework Statement


On high 150km above the surface of Mars, there is satelite That is ciculating arround Mars. What's his bypass time? Radius of Mars is 3400km.

Homework Equations


The Attempt at a Solution



I found equation like this :

t= (2*r(R+h)^(3/2))/(sqrt(g0*R^2))
R=r+h
r=3400km
h=150km

But when i try to calculate like this, i totally miss the actual time. I know bypass time(solution) is 110 minutes.
 
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What value of g0 are you using? What units for r and R are required?
 
welcome to pf!

hi antoman! welcome to pf! :smile:
antoman said:
On high 150km above the surface of Mars, there is satelite That is ciculating arround Mars. What's his bypass time? Radius of Mars is 3400km.

let's put that into english :wink:
At a height of 150km above the surface of Mars, there is a satellite that is orbiting Mars. What's its period? The radius of Mars is 3400km.​

hmm … don't you need to know either the mass of Mars, or the gravitational acceleration (g) at the surface? :redface:
 
SteamKing said:
What value of g0 are you using? What units for r and R are required?

g0=9,81m/s^2

hmm … don't you need to know either the mass of Mars, or the gravitational acceleration (g) at the surface?

i could calculate g at surface using g=g0*R^(2)/(R+h)^2, but then again i have no idea what to use it for
.. nevermind g0 is different on Mars then on Earth so..
 
Ok i made some changes, so now

t= (2*r(R+h)^(3/2))/(sqrt(g0*R^2))
r=R+h
R=3400km
h=150km
g0=3,7 m/s^2... assuming i should of know g0 from mars.

I calculated and it came out t=7261412,315 km/s...:confused:
 
You've got R and h in km and g0 in m/s^2. Don't you think you have a problem with your units?
 
SteamKing said:
You've got R and h in km and g0 in m/s^2. Don't you think you have a problem with your units?


I used 3.6*10^-3 km/s^2 for g0 when i calculated, so its (km*km^(3/2))/sqrt(km*s^(-2)*km^2). Then km^(3/2)/km^(3/2)=1, so all it stays is km*s.
Repair me if I am wrong :)
 
I'm repairing you.
Do you have the right formula?
What if R and r have to be in meters instead of km?
 
SteamKing said:
I'm repairing you.
Do you have the right formula?
What if R and r have to be in meters instead of km?


No, i don't think the formula is right, that's why i asked for help.
 
  • #10
  • #11
Now i know what was the problem :D

Correct formula is:
t=(2*(...))/sqrt(...)
+ g0 to km/s^2, and it all works out perfect, iven makes more sense :)
 

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