Planetary Motion, calculation of orbital period

AI Thread Summary
The discussion centers on calculating the orbital period of a space shuttle orbiting Earth at a specific distance. The gravitational field strength is given as 8.9 N/kg, and the participant is confused about using two different equations for the calculation. It is clarified that while the equation r^3/T^2 is a constant for objects orbiting a particular mass, it is not equivalent to gravitational field strength (g) and lacks the correct dimensions. The correct approach involves using the equation g = 4π²R/T² for accurate results. Understanding the distinction between these formulas is crucial for solving orbital period problems correctly.
Stevo_evo_22
Messages
2
Reaction score
0
Hi everyone,



I'm really confused with a particular question:

Homework Statement






A space shuttle orbits the Earth at 6720 km from its centre. The gravitational field strength is 8.9N/kg. Calculate the shuttle's orbital period in minutes...

Homework Equations



g=4pi^2 x R /T^2

The Attempt at a Solution


The thing is, I know how to solve it using g=4pi^2 x R /T^2, but I want to know why I can't just use g=r^3/T^2 (the constant for all objects orbiting a particular mass)...if I have r and g, why can't i just use this equation?

Thanks!

Steve
 
Physics news on Phys.org
Because r^3/T^2 is a constant. But that constant isn't the same as g. It doesn't even have the right dimensions.
 
Dick said:
Because r^3/T^2 is a constant. But that constant isn't the same as g. It doesn't even have the right dimensions.

ahh ok...thanks :)
 
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...
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...
Back
Top