The velocity of a satellite rotating around the Earth

AI Thread Summary
The discussion centers on the confusion regarding the calculation of a satellite's velocity in relation to its distance from Earth. Participants clarify that the variable 'r' should represent the distance from the Earth's center, not just the altitude. It is noted that if gravity is assumed to be 9.2 m/s², the satellite's distance from the Earth's center would be 6580 km. However, if the altitude is indeed 6580 km, the gravitational acceleration would be approximately 2.38 m/s². Accurate definitions and calculations are essential for understanding satellite motion.
yashboi123
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Homework Statement
A satellite travels at an altitude of 6580 km where gravity is 9.2 m/s^2.
Relevant Equations
v^2/r = a
1694914128479.png

Not sure what r would be in this scenario. I tried adding the radius of the earth to the altitude but that wasn't correct either.
 
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It looks like r is just the 'altitude'.
 
yashboi123 said:
Not sure what r would be in this scenario. I tried adding the radius of the earth to the altitude but that wasn't correct either.
Please show your work. We cannot help you if you don't tell us what you did.
 
The confusion is in the question itself.

If gravity is 9.2 m/s^2 (as stated in the question), then satellite distance from earth center is 6580km (verify yourself!). And that distance should not be called altitude :-(

If altitude is really 6580km, g is about 2.38 m/s^2.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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