Gravity acceleration 1/5 earths, how high?

AI Thread Summary
To determine how high above Earth's surface gravity is one fifth of its surface value, one must consider the inverse square law of gravitation, which states that gravitational force decreases with the square of the distance from the center of the Earth. The acceleration due to gravity at a distance 'r' from the center can be expressed as g' = g / (r^2/R^2), where g is the surface gravity and R is the Earth's radius. Setting g' to one fifth of g allows for the calculation of the required distance from the center of the Earth. The problem emphasizes the need for initial thoughts and calculations to facilitate assistance. Understanding gravitational changes with distance is crucial for solving this problem effectively.
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Homework Statement



How far above the Earth's surface will the acceleration of gravity be one fifth of what it is on the surface?
 
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We can only help those who help themselves. What thoughts have you had on the problem already? What work have you already done? Without this information, we cannot (and will not) help you.
 
Hoe does gravity falls off with distance ? Remember that distances are measured from the centre of the Earth
 
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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