Physics acceleration due to gravity

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The gravitational force on a spacecraft at a distance of 4R from Earth's center is calculated using Newton's law of universal gravitation, resulting in F = G(Mm)/(16R^2). The weight of the spacecraft at this distance is W = mg, where g is the gravitational acceleration at 4R, which is one-sixteenth of that at Earth's surface. The weight can also be expressed in terms of the acceleration due to gravity at 4R as W = m(g/16). The gravitational acceleration at 4R can be expressed as g(4R) = GM/(16R^2), where G is the gravitational constant and M is Earth's mass. Understanding these relationships is crucial for spacecraft dynamics and orbital mechanics.
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A spacecraft of mass m at a distance of 4R(earth) from the center of the planet. a) What is the formula for the gravitational force on the vehicle at that distance? b) What is the weight of the vehicle at a distance of 4R(earth)? c) Alternatively, express the weight in terms of the acceleration due to gravity at 4R(earth). d) Write an expression for the gravitational acceleration at 4R, in terms of R(earth), G, and M(earth).
 
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jjbrin00 said:
A spacecraft of mass m at a distance of 4R(earth) from the center of the planet. a) What is the formula for the gravitational force on the vehicle at that distance? b) What is the weight of the vehicle at a distance of 4R(earth)? c) Alternatively, express the weight in terms of the acceleration due to gravity at 4R(earth). d) Write an expression for the gravitational acceleration at 4R, in terms of R(earth), G, and M(earth).

Have a look at this. Especially the summaries in the two blue boxes.

http://csep10.phys.utk.edu/astr161/lect/history/Newtongrav.html
 
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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