Gravitational Self Energy of a Sphere

Whenever a stable system is formed, it loses some of its energy. Similarly, to break it and separate its entities to infinity we must give it at least the same amount of energy. This energy is called the self-potential energy of the system. Technically, self-potential energy is defined as the amount of work done by the internal forces in assembling a system from infinity or work done by an external agent to slowly break the system and separate its entities to infinity.

The gravitational self-potential energy of a solid sphere of mass density ρ \rho and radius R R is E E . What is the gravitational self-potential energy of a sphere of mass density 2 ρ 2\rho and radius 2 R 2R ?

4 E 4E 8 E 8E 16 E 16E 32 E 32E 128 E 128E None of the rest

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2 solutions

Michael Huang
Jan 11, 2017

Relevant wiki: Evaluating Gravitational Potential Energy

The formula of the gravitational potential energy is E = 3 G M 2 5 R . E = \dfrac{3GM^2}{5R}. where M M is the mass, R R is the radius and G G is the gravitational constant. Looking at the following equation M = 4 3 π R 3 ρ . M = \dfrac{4}{3}\pi R^3 \rho. since both R R and ρ \rho are doubled, then M doubled = 4 3 π ( 2 R ) 3 ( 2 ρ ) = 16 ( 4 3 R 3 ρ ) = 16 M . M_{\text{doubled}} = \dfrac{4}{3} \pi (2R)^3 (2\rho) = 16 \left(\dfrac{4}{3} R^3 \rho\right) = 16 M. Thus, E doubled = 3 G ( 16 M ) 2 5 ( 2 R ) . = 1 6 2 2 3 G M 2 5 R . = 128 E . \begin{array}{rl} E_{\text{doubled}} &= \dfrac{3G\left(16M \right)^2}{5(2R)}.\\ &= \dfrac{16^2}{2} \cdot \dfrac{3GM^2}{5R}.\\ &= \boxed{128E}. \end{array}


Relationship Between E E , M M and R R


If we were to generalize this for arbitrary constants k 1 k_1 and k 2 k_2 for ρ n = k 1 ρ \rho_n = k_1\rho and R n = k 2 R R_n = k_2 R (where k 1 , k 2 > 0 k_1, k_2 > 0 ), and E E is the original potential energy, then the new potential energy is E general = 3 G 5 ( 4 3 π ( k 2 ) 3 R 3 k 1 ρ ) 2 k 2 R . = 3 G 5 ( k 2 ) 6 ( k 1 ) 2 ( 4 3 π R 3 ρ ) 2 k 2 R . = ( k 1 ) 2 ( k 2 ) 5 3 G M 2 5 R . = ( k 1 ) 2 ( k 2 ) 5 E . \begin{array}{rl} E_{\text{general}} &= \dfrac{3G}{5} \cdot \dfrac{\left(\frac{4}{3} \pi \left(k_2\right)^3 R^3 k_1 \rho\right)^2}{k_2 R}.\\ &= \dfrac{3G}{5} \cdot \dfrac{\left(k_2\right)^6\left(k_1\right)^2 \left(\frac{4}{3} \pi R^3 \rho\right)^2}{k_2 R}.\\ &= \left(k_1\right)^2\left(k_2\right)^5 \cdot \dfrac{3GM^2}{5R}.\\ &= \left(k_1\right)^2\left(k_2\right)^5 E. \end{array} So E E depends on the constants k 1 k_1 and k 2 k_2 .

Shouldn't it be k 1 2 k_1^2 in your generalisation?

Sumanth R Hegde - 4 years, 5 months ago

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Typo fixed. Thanks for the catch!

Michael Huang - 4 years, 5 months ago

This problem can be solved through dimensional analysis also, without even knowing the formula.

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