Question

Mass M is distributed uniformly over a rod of length $L$. Find the gravitational field at P at a distance a on perpendicular bisector.

• A. $\frac{2GM}{a(L^2+a^2)}$
• B. $\frac{2GM}{a(L^2+4a^2{)}^{1/2}}$
• C. $\frac{GM}{2a(L^2+4a^2)}$
• D. none

Answer 1

Answer: (B)

$\frac{2GM}{a(L^2+4a^2{)}^{1/2}}$

$dm=\frac{M}{L}dx$

$dE=\frac{G\frac{M}{L}dx}{(x^2+a^2)}d{E}_{net}=dEcos\theta =\frac{GMadx}{L(x^2+a^2{)}^{3/2}}$

$E=\int dEcos\theta =\frac{GMa}{L}{\int }_{-L/2}^{L/2}\frac{dx}{(x^2+a^2{)}^{3/2}}$

Put $x=atan\theta$ then $dx=ase{c}^2\theta d\theta$

$E=\frac{GMa}{L}\int \frac{ase{c}^2\theta d\theta }{a^{3}se{c}^3\theta }=\frac{GM}{La}\int cos\theta d\theta$

$=\frac{GM}{La}sin\theta =\frac{GMx}{La\sqrt{x^2+a^2}}{\int }_{-L/2}^{L/2}$

$=\frac{GM\left(2L\right)}{aL(L^2+4a^2{)}^{1/2}}=\frac{2GM}{a(L^2+4a^2{)}^{1/2}}$