Calculating Weight in Orbit: How Does Distance Affect Gravitational Force?

AI Thread Summary
To calculate the weight of a mass in orbit 3000 km above a planet with a 3000 km radius, the gravitational force equation F = (Gm1m2)/r² is relevant. The original weight of 20 N at the planet's surface indicates the gravitational acceleration, which can be used to find the mass. The new radius for the orbit is the sum of the planet's radius and the altitude, totaling 6000 km. Understanding how to adjust the radius in the gravitational formula is crucial for determining the new weight in orbit. The discussion emphasizes the need to clarify the relationship between the surface radius and the orbital radius for accurate calculations.
abpandanguyen
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Homework Statement



A mass weighs 20 N at the surface of a planet of 3000 km radius. What is the mass's weight when it is in orbit 3000 km above the surface of the planet?

Homework Equations


F = (Gm1m2)/r2 (well... not really)
Ug = mgy


The Attempt at a Solution


I don't even know. I'm trying to find the g on this planet but I'm not sure where that would lead me... Help please >_<
 
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what is the new 'r' you want to find? How does that relate to the original 'r'?

r= r at surface then

GMm/r2= ?
 
yeah, i just needed to write the equations out instead of just thinking in my head

thanks!
 
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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