Final answer:
The equation for the density of Earth is dimensionally consistent, and using the Earth's mass and radius allows for the verification of the gravitational constant's accepted value of approximately 6.674×10-11 N·m²/kg².
Step-by-step explanation:
To check the dimensional consistency of the given expression
P= 3g/4πRG, where R is the radius of the Earth, G is the gravitational constant, and g is the acceleration due to gravity, we'll examine the dimensions of each term.
Let's assign dimensions as follows:
P (density) has dimensions ML−3 (mass per unit volume).
g (acceleration due to gravity) has dimensions LT−2 (length per unit time squared).
R (radius of the Earth) has dimensions L (length).
G (gravitational constant) has dimensions L3M−1T−2(cubed length times mass to the power of negative one, times time to the power of negative two).
Now, let's substitute these dimensions into the given expression:
[P]= 3[ML−3][LT−2]/4π[L][L3M−1T−2]
Now, simplify the expression:
[P]= 3/4π. [M][L−3][LT−2]/ [L][L3M−1T−2 ]
[P]= 3/4π. [M][L−2][T−2]/ [L2][M−1]
[P]= 3/4π. [M][T−2]/ [L][M−1]
Now, rearrange the terms to group like dimensions:
[P]= 3/4π. [M][M−1 ][T−2]/ [L]
[P]= 3/4π. [T−2]/[L]
So, the expression is dimensionally consistent, as both sides of the equation have the same dimensions (ML−3).