What happens to the height of the orbit?

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A satellite in circular orbit around the sun requires a greater change in speed to escape the solar system than to fall into the sun, as the escape velocity is significantly higher. The initial speed needed to escape to infinity is greater than that required to reach the sun due to gravitational dynamics. Atmospheric friction in low Earth orbit increases a satellite's speed because it creates drag that can alter the satellite's trajectory and energy. This friction can lead to a decrease in altitude, affecting the satellite's orbital path. Understanding these dynamics is crucial for satellite mission planning and orbital mechanics.
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Homework Statement



1) A satellite is in circular orbit around the sun. Which requires the greater change in satellite's speed: to escape the solar system or fall into the sun?

2)Why does atmospheric friction increase the speed of a satellite in low Earth orbit?

Homework Equations





The Attempt at a Solution



1) Not sure..Surely the change in speed in 2 cases would be the same?

2) Again ..not sure.. :S
 
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hi bon! :smile:
bon said:
1) A satellite is in circular orbit around the sun. Which requires the greater change in satellite's speed: to escape the solar system or fall into the sun?

1) Not sure..Surely the change in speed in 2 cases would be the same?

(if the Sun has mass M and if the orbit has radius r …)

What initial speed is needed to escape to infinity?

What initial speed is needed to reach the Sun?
2)Why does atmospheric friction increase the speed of a satellite in low Earth orbit?

It stays in orbit: but what happens to the orbit?
 
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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