Newton's law of universal gravitation states that the gravitational force F between two po… — JAMB UTME

Newton's law of universal gravitation states that the gravitational force $F$ between two point masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = \frac{G m_1 m_2}{r^2}$ , where $G$ is the universal gravitational constant. What is the dimensional formula of $G$ ?

A
$M L^3 T^{-2}$
B
$M^{-1} L^2 T^{-2}$
$M^{-1} L^3 T^{-2}$ Cevap
D
$M L^{-1} T^{-2}$

Cevap

The dimensional formula of the universal gravitational constant

G

M^{-1} L^3 T^{-2}

Rearranging

F = \frac{G m_1 m_2}{r^2}

gives

G = \frac{F r^2}{m_1 m_2}

. Substituting the dimensions of Force (

M L T^{-2}

), distance squared (

L^2

), and mass squared (

M^2

) gives

\frac{(M L T^{-2})(L^2)}{M^2} = M^{-1} L^3 T^{-2}

Adım Adım Çözüm

Rearrange the gravitational force equation to solve for

G

G = \frac{F r^2}{m_1 m_2}

Isolating the physical constant allows us to substitute the dimensions of each constituent quantity.

Substitute the fundamental dimensions for force, distance, and mass.

[F] = M L T^{-2}

[r] = L

[m_1] = [m_2] = M

Force is mass times acceleration (

M \cdot L T^{-2}

), distance is length (

L

), and mass is

[M]

Substitute these fundamental dimensions into the expression for

G

and simplify the exponents.

[G] = \frac{(M L T^{-2}) (L^2)}{M^2} = M^{1-2} L^{1+2} T^{-2} = M^{-1} L^3 T^{-2}

Applying standard exponent rules yields the final dimensional formula.

Anahtar Kavram

Deriving Dimensions of Physical Constants

Soru

Cevap

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