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Question 2.56: All of electrostatics follows from the 1/r^ 2 character of C...

All of electrostatics follows from the 1/r ^2 character of Coulomb’s law, together with the principle of superposition. An analogous theory can therefore be constructed for Newton’s law of universal gravitation. What is the gravitational energy of a sphere, of mass M and radius R, assuming the density is uniform? Use your result to estimate the gravitational energy of the sun (look up the relevant numbers). Note that the energy is negative—masses attract, whereas (like) electric charges repel. As the matter “falls in,” to create the sun, its energy is converted into other forms (typically thermal), and it is subsequently released in the form of radiation. The sun radiates at a rate of 3.86 \times 10^{26} W; if all this came from gravitational energy, how long would the sun last? [The sun is in fact much older than that, so evidently this is not the source of its power.^{14} ]

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Compare Newton’s law of universal gravitation to Coulomb’s law:

F =-G \frac{m_{1} m_{2}}{r^{2}} \hat{ r } ; \quad F =\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} \hat{ r } .

Evidently \frac{1}{4 \pi \epsilon_{0}} \rightarrow G \text { and } q \rightarrow m . The gravitational energy of a sphere (translating Prob. 2.34) is therefore

W_{ grav }=\frac{3}{5} G \frac{M^{2}}{R} .

Now, G=6.67 \times 10^{-11} N m ^{2} / kg ^{2} , and for the sun M=1.99 \times 10^{30} kg , R=6.96 \times 10^{8} m , so the sun’s gravitational energy is W=2.28 \times 10^{41} J . At the current rate this energy would be dissipated in a time

t=\frac{W}{P}=\frac{2.28 \times 10^{41}}{3.86 \times 10^{26}}=5.90 \times 10^{14} s =1.87 \times 10^{7} \text { years. }

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