Derive the dimension of the Universal Gravitational Constant (G).

March 30, 2026Physics

ScanSolve AI Answer

March 30, 2026

Answer

\frac{Fr^2}{m_1 m_2}$

Here are the solutions to the dimensional analysis problems.

1. Derived the dimension of the following physical quantities:

Universal Gravitational Constant ( $G$ ) Formula: $G = \frac{Fr^2}{m_1 m_2}$ Dimensions: $[F] = M L T^{-2}$ $[r^2] = L^2$ $[m_1 m_2] = M^2$

Was this helpful?

Need help with your own homework?

Get instant step-by-step solutions to any question. Free to start.

Ask Your Question

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp

More Physics Questions

QWhy cloud index model is suitable for optimize the amount of insolation using satellite data QMore than 20 word problems In kinetic energy QWhy optical weight is attached to angstrom model for optimization QWhy optical weight is suitable for the optimization QExplain why angstrom model is suitable for optimize ground based insolation

See all Physics questions →

Still have questions?

Quick Answer

Universal Gravitational Constant (G) Formula: G = (Fr^2)/(m_1 m_2) Dimensions: [F] = M L T^-2 [r^2] = L^2 [m_1 m_2] = M^2 [G] = M L T^-2 · L^2M^2 = M L^3 T^-2M^2…

Asked on March 30, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

ScanSolve AI AnswerAnswered on March 30, 2026

Here are the solutions to the dimensional analysis problems. 1. Derived the dimension of the following physical quantities: Universal Gravitational Constant (G) Formula: G = (Fr^2)/(m_1 m_2) Dimensions: [F] = M L T^-2 [r^2] = L^2 [m_1 m_2] = M^2 [G] = M L T^-2 · L^2M^2 = M L^3 T^-2M^2