Kansas State University

1. $5.97 \times 10^{24} kg$.

3. $1.899 \times 10^{27} kg$.

5. $8.49 \times 10^{11} m$.

7. $1.3 \times 10^{16} m$.

9. We have perihelion $1/r_p = \mu/c^2 + A$ and aphelion $1/r_a=\mu/c^2 - A$ where $A$ is the semi-major axis. Adding these together we get $$\begin{align} \frac{2\mu}{c^2} &= \frac{1}{r_p} + \frac{1}{r_a} \\ c^2 &= \frac{2\mu}{1/r_p + 1/r_a} \\ c &= \sqrt{\mu\frac{2r_pr_a}{r_p+r_a}} \end{align}$$ Then since velocity at perihelion and aphelion is $c/r$, we divide by $r_p$ to get velocity at perihelion and by $r_a$ to get velocity at aphelion, which give the desired formulas.

11. Approximately 54,500 m/sec.

13. $1.24 \times 10^{10} sec$ (which is about 393 years).

15. Approximately 69,500 m/sec.

17. Approximately $32,500 m/sec$.

19. Since the transverse component of acceleration is still 0, the proof of the second law in the text works exactly the same for Bennett's law of gravity as it does for Newton's.

21. Our key equation is $$ r'' - r(\theta')^2 = - \frac{GM}{r^n}. $$ Using the conservation of angular momentum, $\displaystyle \theta' = \frac{c}{r^2}$, we get $$ \begin{align} r'' - r(\theta')^2 &= -\frac{GM}{r^n} \\ r'' - r\left(\frac{c}{r^2}\right)^2 &= -\frac{GM}{r^n} \\ r'' - \left(\frac{c^2}{r^3}\right) &= -\frac{GM}{r^n} \\ -\frac{r^2r''}{c^2} + \frac{1}{r} &= \left(\frac{GM}{c^2}\right)r^{2-n} \end{align} $$ Then making the substitution $u=1/r$, which leads to $\displaystyle \frac{d^2u}{d\theta^2} = -\frac{r^2r''}{c^2}$ as shown in the text we get $$ \frac{d^2u}{d\theta^2} + u = \left(\frac{GM}{c^2}\right)u^{n-2}. $$ For this to be a linear equation, we can only have $u$ to the zeroth or first power, so we only get a linear equation if $n=2$ or $n=3$.

23. We want points $(x,y)$ where the sum of the distances to $(f,0)$ and $(-f,0)$ is $2a$. From the distance formula we thus get $$ \sqrt{(x-f)^2+y^2} + \sqrt{(x+f)^2+y^2} = 2a. $$ To simplify this equation we isolate one of the square roots and square both sides, then isolate the remaining square root and square both sides again as follows. $$ \begin{align} \sqrt{(x-f)^2+y^2} + \sqrt{(x+f)^2+y^2} &= 2a \\ \sqrt{(x-f)^2+y^2} &= 2a - \sqrt{(x+f)^2+y^2} \\ (x-f)^2+y^2 &= 4a^2 - 4a\sqrt{(x+f)^2+y^2} + (x+f)^2 + y^2 \\ x^2 - 2xf + f^2 +y^2 &= 4a^2 - 4a\sqrt{(x+f)^2+y^2} + x^2+2xf+f^2 + y^2 \\ -2xf &= 4a^2 - 4a\sqrt{(x+f)^2+y^2} + 2xf \\ 4a\sqrt{(x+f)^2+y^2} &= 4a^2+4xf \\ a\sqrt{(x+f)^2+y^2} &= a^2+xf \\ a^2\left((x+f)^2 + y^2\right) &= a^4 +2a^2xf + x^2f^2 \\ a^2x^2 + 2a^2xf + a^2f^2 + a^2y^2 &= a^2 + 2a^2xf + x^2f^2 \\ (a^2-f^2)x^2 + a^2y^2 &= a^4-a^2f^2 \\ (a^2-f^2)x^2 + a^2y^2 &= a^2(a^2-f^2) \\ \left(\frac{x^2}{a^2}\right) + \left(\frac{y^2}{a^2 - f^2}\right) &=1. \end{align} $$ 25. We compute $$ \begin{align} b^2 + f^2 &= \frac{e^2p^2}{1-e^2} + \frac{e^4p^2}{(1-e^2)^2} \\ &= \frac{(1-e^2)e^2p^2}{(1-e^2)^2} + \frac{e^4p^2}{(1-e^2)^2} \\ &= \frac{e^2p^2 - e^4p^2 + e^4p^2}{(1-e^2)^2} \\ &= \frac{e^2p^2}{(1-e^2)^2} \\ &= \left(\frac{ep}{1-e^2}\right)^2 \\ &= a^2. \end{align} $$