# [SOLVED] gravitational attraction

#### stealth

Hi:
a) Two lead balls with masses 5.20kg and 0.250kg are place with their centres 50.0cm apart. With what gravitational force do they attract each other?
b) At Earth's surface g=9.806m/s². Assuming the Earth to be a sphere of radius 6.371x10^6, find the mass of the Earth.
Take G = 6.67x10^-11Nm²kg^-2

a) Use: G(Mb1)(Mb2)/r^2
(6.67x10^-11)(5.20)(0.250)/(0.5)=3.5x10^-10 (2s.f.)

b) Use: g=GM/r^2
9.806x(6.371x10^6)^2/6.67x10^-11=6.0x10^24 (2s.f.)

Hopefully my calculations are good. Please check if ok someone?(Nerd)

#### arbolis

PHF Hall of Fame
Hi:
a) Two lead balls with masses 5.20kg and 0.250kg are place with their centres 50.0cm apart. With what gravitational force do they attract each other?
b) At Earth's surface g=9.806m/s². Assuming the Earth to be a sphere of radius 6.371x10^6, find the mass of the Earth.
Take G = 6.67x10^-11Nm²kg^-2

a) Use: G(Mb1)(Mb2)/r^2
(6.67x10^-11)(5.20)(0.250)/(0.5)=3.5x10^-10 (2s.f.)

b) Use: g=GM/r^2
9.806x(6.371x10^6)^2/6.67x10^-11=6.0x10^24 (2s.f.)

Hopefully my calculations are good. Please check if ok someone?(Nerd)
Welcome to PHF!
b) looks good to me. But a), I'm not sure if it's a typo or an error. You forgot to square the distance between both balls.

#### stealth

ahh! I did too(Shake)

a) Use: G(Mb1)(Mb2)/r^2
(6.67x10^-11)(5.20)(0.250)/(0.5)²=3.5x10^-10 (2s.f.)

Is that what you mean?(Thinking)

#### arbolis

PHF Hall of Fame
ahh! I did too(Shake)

a) Use: G(Mb1)(Mb2)/r^2
(6.67x10^-11)(5.20)(0.250)/(0.5)²=3.5x10^-10 (2s.f.)

Is that what you mean?(Thinking)
Yes, that's what I meant.

#### stealth

great! Excuse the error

Thanks for you help arbolis (Cool)