Math  /  Calculus

QuestionProblems 12.82 and 12.83 Two masses mAm_{A} and mBm_{B} are placed at a distance r0r_{0} from one another. Because of their mutual gravitational attraction, the acceleration of sphere BB as seen from sphere AA is given by r¨=G(mA+mBr2)\ddot{r}=-G\left(\frac{m_{A}+m_{B}}{r^{2}}\right) where G=6.674×1011 m3/(kgs2)=3.439×108ft3/(G=6.674 \times 10^{-11} \mathrm{~m}^{3} /\left(\mathrm{kg} \cdot \mathrm{s}^{2}\right)=3.439 \times 10^{-8} \mathrm{ft}^{3} /\left(\right. slug s2)\left.\cdot \mathrm{s}^{2}\right) is the universal gravitational constant. Problem 12.82 If the spheres are released from rest, determine (a) The velocity of BB (as seen by AA ) as a function of the distance rr. (b) The velocity of BB (as seen by AA ) at impact if r0=7ftr_{0}=7 \mathrm{ft}, the weight of AA is 2.1 lb , the weight of BB is 0.7 lb , and (i) The diameters of AA and BB are dA=1.5ftd_{A}=1.5 \mathrm{ft} and dB=1.2ftd_{B}=1.2 \mathrm{ft}, respectively. (ii) The diameters of AA and BB are infinitesimally small. Answer (a) r˙=2G(mA+mB)r0rr0\dot{r}=-\sqrt{2 G\left(m_{A}+m_{B}\right)} \sqrt{\frac{r_{0}-r}{r_{0}}}; (b) (i) r˙=5.98.0×105ft/s\dot{r}=-5.98 .0 \times 10^{-5} \mathrm{ft} / \mathrm{s}, (ii) r˙\dot{r} \rightarrow-\infty
Problem 12.83 Assume that the particles are released from rest at r=r0r=r_{0}. (a) Determine the expression relating their relative position rr and time. Hint: x/(1x)dx=sin1(x)x(1x)\int \sqrt{x /(1-x)} d x=\sin ^{-1}(\sqrt{x})-\sqrt{x(1-x)} (b) Determine the time it takes for the objects to come into contact if r0=3 m,Ar_{0}=3 \mathrm{~m}, A and BB have masses of 1.1 and 2.3 kg , respectively, and (i) The diameters of AA and BB are dA=22 cmd_{A}=22 \mathrm{~cm} and dB=15 cmd_{B}=15 \mathrm{~cm}, respectively. (ii) The diameters of AA and BB are infinitesimally small.

Studdy Solution
For the first contact (n=1n = 1): t=π2.244×1061.4×106 s16.2 days t = \frac{\pi}{2.244 \times 10^{-6}} \approx 1.4 \times 10^6 \ \text{s} \approx 16.2 \ \text{days}
_SOLUTION_: (a) r˙=2G(mA+mB)(r0rr0)\dot{r} = -\sqrt{2 G (m_A + m_B) \left( \frac{r_0 - r}{r_0} \right)} (b) (i) r˙5.98×105 ft/s\dot{r} \approx -5.98 \times 10^{-5} \ \text{ft/s} (ii) r˙\dot{r} \rightarrow -\infty (c) r=r0sin2(π22G(mA+mB)r0t)r = r_0 \sin^2 \left( \frac{\pi}{2} - \frac{\sqrt{2 G (m_A + m_B)}}{r_0} t \right) (d) (i) t8.11 dayst \approx 8.11 \ \text{days} (ii) t16.2 dayst \approx 16.2 \ \text{days}
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