The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

gravitational field strength

So gravitational field strength is defined as the force exerted per unit mass at a point in the field. So g=F/M so then in my textbook it then goes onto say that it's independent of the test mass used, but how can this be? from the equation surley a leager test mass means a smaller g field so it's dependent?
bree crazy gravitons with infinite range cause they have no mass but are affected by mass you get me ?
Reply 2
Avatar for Benniboi1
Benniboi1
9
Original post by Emissionspectra
So gravitational field strength is defined as the force exerted per unit mass at a point in the field. So g=F/M so then in my textbook it then goes onto say that it's independent of the test mass used, but how can this be? from the equation surley a leager test mass means a smaller g field so it's dependent?


doesn't the M in g=F/M represent the Mass of the thing creating the gravitational force though, eg M would be the earth then a larger test mass wont affect it
Original post by Benniboi1
doesn't the M in g=F/M represent the Mass of the thing creating the gravitational force though, eg M would be the earth then a larger test mass wont affect it


thinking about it i think its becuase Force has mass in its units so the m's cancel out.
Reply 4
Avatar for teachercol
teachercol
9
Grav force on a mass is prop to the mass.

F=GMm/r^2

g = F/m
It's independent because the "F" in your equation is the gravitational force on the mass (which is proportional to the smaller mass in the field) and so the two masses cancel.


g=Fgravitationalmg=\frac{F_{gravitational}}{m}, where m is the mass in the gravitational field

Fgravitational=GMmr2F_{gravitational}=\frac{GMm}{r^{2}}, where M is the mass causing the gravitational field and m is the mass in the gravitational field.

Combining these two, we get g=GMmr2mg=\frac{\frac{GMm}{r^{2}}}{m}, and so both the lowercase ms cancel out, giving g=GMr2g=\frac{GM}{r^{2}}. From this we can see that g is proportional only to the mass causing the field and the square of the distance from that mass; it is not proportional to the mass experiencing the field.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you