Calculate Moon's Gravity Acceleration | Radius 1.74x10^6m | Mass 7.35x10^22kg

Thread starter Mazdak
Start date Oct 28, 2004
Tags

Acceleration

AI Thread Summary

To calculate the acceleration due to gravity on the Moon, Newton's law of gravity and Newton's second law are essential. The Moon's radius is approximately 1.74x10^6 meters, and its mass is 7.35x10^22 kilograms. The formula for gravitational acceleration is derived from these principles. Understanding these laws is crucial for solving the problem effectively. The discussion emphasizes the importance of applying these concepts to find the solution.

Oct 28, 2004

Mazdak

Calculate the acceleration due to gravity on the Moon. The Moon's radius is about 1.74x10^6m and its mass is 7.35x10^22kg

Heeeeelp ! PLEASE It's urgent

Physics news on Phys.org

Oct 28, 2004

Fredrik

Staff Emeritus

Science Advisor

Homework Helper

Insights Author

Gold Member

Mazdak, you should tell us about your own ideas on how to solve this problem. Do you know Newton's law of gravity and Newton's second law? You will need to use both.

Oct 28, 2004

Mazdak

aha...ok thanks, now I understand

Thread 'I've come across another fluid pressure problem I don't understand'

Neglect gravity other than that of water; neglect friction. F1=ρgsh=1000*10*1*(1+4)=50000 N; F1=ρgsh=1000*10*0.1*4=4000 N; F3=50000-4000=46000 N?

View full post »

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}...

View full post »

Thread 'Understanding Forces and their Vector Components'

I was told forces are vectors so they can be divided into x and y components, but what about the normal force or the force of static friction? do you divide those into x and y components if say you had a block that was lying on an angled surface?

View full post »