How High Must a Satellite Orbit to Remain Geostationary Over Jupiter?

AI Thread Summary
To determine the altitude for a geostationary satellite over Jupiter, the gravitational force must equal the centripetal force acting on the satellite. Jupiter's rotation period is 9.84 hours, and its mass is approximately 1.90E27 kg with a mean radius of 6.99E7 m. The calculations involve equating gravitational force to radial acceleration, leading to the formula F_grav = m(v^2/r). It's noted that the satellite must be positioned above the equator for true geostationarity, although the Great Red Spot is not located there. The discussion emphasizes the need for understanding the relationship between gravitational and centripetal forces to solve the problem.
ookt2c
Messages
16
Reaction score
0

Homework Statement


A synchronous satellite, which always remains above the same point on a planet's wquator, is put in orbit around Jupiter to study the famous red spot. Jupiter rotates once every 9.84 h. find the altitude of the satelite


Homework Equations



jupiter: mass in kg 1.90E27,Mean radius 6.99E7,period of revolutions3.74E8,Mean distance from the sun7.78E11 (t^2/R^3)(S^2/M^3)= 2.97E-19

The Attempt at a Solution



I don't know where to start , missed a day of class
 
Physics news on Phys.org
The force of gravity must be equal to the centripetal force. (But this is true for even if the satellite is not synchronous.)
 
ookt2c said:

Homework Statement


A synchronous satellite, which always remains above the same point on a planet's wquator, is put in orbit around Jupiter to study the famous red spot. Jupiter rotates once every 9.84 h. find the altitude of the satelite


Homework Equations



jupiter: mass in kg 1.90E27,Mean radius 6.99E7,period of revolutions3.74E8,Mean distance from the sun7.78E11 (t^2/R^3)(S^2/M^3)= 2.97E-19

The Attempt at a Solution



I don't know where to start , missed a day of class

Strictly speaking, this is possible only if the point on th eplanet is at the equator. I don't think that the Red Spot is on th eequator but let's pretend it is.

Then you have to impose that the force of gravity on the satellite is equal to m tiomes the radial acceleration so
F_{grav} = m \frac{v^2}{r}

where "r" is the distance from the satellite to the center of the planet.
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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