Calculating the Force of Gravity on a 1-kg Mass at Twice Earth's Radius

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The force of gravity on a 1-kg mass at twice Earth's radius (1.28 x 10^7 m) is calculated using the formula F = G(M1*M2)/r^2. Substituting the known values, the calculation yields a gravitational force of 2.4 N. This force represents the attraction between the 1-kg mass and Earth at that distance. The discussion confirms the accuracy of the calculation and clarifies that it pertains to gravitational attraction. Understanding this concept is essential for grasping gravitational forces at varying distances from Earth.
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Calculate the force of gravity on a 1-kg mass if it were 6.4 x 10^6 m above Earth's surface (twice Earth's radius).

Known: Mass of object 1 = 1 kg
Mass of Earth = 6 x 10^24 kg
Earth's radius = 6.4 x 10^6 m; 2x this is 1.28 x 10^7 m
G = 6.67 x 10^-11 N*m^2/kg^2

Equation: F = G(Mass of object)(Mass of Earth)/(Distance)^2
= (6.67 x 10^-11 N*m^2/kg^2)(1 kg)(6 x 10^24 kg)/(1.28 x10 ^7 m)^2
= 2.4 N
 
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yes you did it right

it's simply asking you for hte attraction force between the object and earth, really.
 
Ok, thank you for checking.
 
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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