Question

gravitational field due to finite wire

Answer 1

Detailed calculation involving integration of gravitational field contributions from infinitesimal mass elements of the wire. The field depends on the position of the point relative to the wire. For a point at perpendicular distance r from a wire of length L, the field components can be calculated by integrating along the length of the wire. Specific formulas are obtained for points opposite the center or opposite an end of the wire.

Answer 2

The gravitational field due to a finite wire of length L and mass M can be calculated at a point P using integration. Let the wire lie along the x-axis. Consider a point P at a perpendicular distance r from the wire. Let the wire extend from $x_1$ to $x_2$ along the x-axis, such that $L = x_2 - x_1$. The linear mass density is $\lambda = M/L$.

Consider a small element of the wire of length dx at position (x, 0). The mass of this element is $dm = \lambda dx$. The distance from this element to the point P at (0, r) is $R = \sqrt{x^2 + r^2}$. The gravitational field dE due to this element at P is $dE = \frac{G dm}{R^2} = \frac{G \lambda dx}{x^2 + r^2}$, directed from P towards the element (x, 0).

Let $\theta$ be the angle the line connecting P and the element makes with the perpendicular line from P to the wire (the y-axis in this setup). The vector from P(0, r) to (x, 0) is $(x, -r)$. The angle $\theta$ can be defined such that $\tan \theta = x/r$. The components of dE along the x and y directions are: $dE_x = dE \sin \theta$ (towards the wire if x is positive, away if x is negative) $dE_y = dE \cos \theta$ (towards the wire, i.e., downwards in this setup)

From the geometry, $\sin \theta = \frac{x}{\sqrt{x^2 + r^2}}$ and $\cos \theta = \frac{r}{\sqrt{x^2 + r^2}}$. So, $dE_x = \frac{G \lambda dx}{x^2 + r^2} \frac{x}{\sqrt{x^2 + r^2}} = \frac{G \lambda x dx}{(x^2 + r^2)^{3/2}}$. And $dE_y = \frac{G \lambda dx}{x^2 + r^2} \frac{r}{\sqrt{x^2 + r^2}} = \frac{G \lambda r dx}{(x^2 + r^2)^{3/2}}$.

To find the total gravitational field components $E_x$ and $E_y$ at P, we integrate these expressions from $x_1$ to $x_2$: $E_x = \int_{x_1}^{x_2} \frac{G \lambda x dx}{(x^2 + r^2)^{3/2}} = G \lambda \left[ \frac{-1}{\sqrt{x^2 + r^2}} \right]_{x_1}^{x_2} = -G \lambda \left( \frac{1}{\sqrt{x_2^2 + r^2}} - \frac{1}{\sqrt{x_1^2 + r^2}} \right) = G \lambda \left( \frac{1}{\sqrt{x_1^2 + r^2}} - \frac{1}{\sqrt{x_2^2 + r^2}} \right)$. $E_y = \int_{x_1}^{x_2} \frac{G \lambda r dx}{(x^2 + r^2)^{3/2}}$. Let $x = r \tan \phi$, so $dx = r \sec^2 \phi d\phi$. When $x=x_1$, $\tan \phi_1 = x_1/r$. When $x=x_2$, $\tan \phi_2 = x_2/r$. $x^2 + r^2 = r^2 \tan^2 \phi + r^2 = r^2 \sec^2 \phi$. $(x^2 + r^2)^{3/2} = r^3 \sec^3 \phi$. $E_y = \int_{\phi_1}^{\phi_2} \frac{G \lambda r (r \sec^2 \phi d\phi)}{r^3 \sec^3 \phi} = \frac{G \lambda}{r} \int_{\phi_1}^{\phi_2} \frac{\sec^2 \phi}{\sec^3 \phi} d\phi = \frac{G \lambda}{r} \int_{\phi_1}^{\phi_2} \cos \phi d\phi = \frac{G \lambda}{r} [\sin \phi]_{\phi_1}^{\phi_2} = \frac{G \lambda}{r} (\sin \phi_2 - \sin \phi_1)$. Here, $\phi_1$ and $\phi_2$ are the angles such that $\tan \phi_1 = x_1/r$ and $\tan \phi_2 = x_2/r$. Alternatively, we can use the angles the ends of the wire make with the perpendicular line from P. Let $\alpha_1$ be the angle made by the line from P to $x_1$ with the perpendicular, and $\alpha_2$ be the angle made by the line from P to $x_2$ with the perpendicular. If the wire is from $x=-a$ to $x=b$, then $x_1=-a$ and $x_2=b$. The angles $\phi_1$ and $\phi_2$ are such that $\tan \phi_1 = -a/r$ and $\tan \phi_2 = b/r$. The angle $\alpha_1$ made by the line to $x_1=-a$ with the perpendicular (y-axis) is such that $\tan \alpha_1 = |-a|/r = a/r$. Let's measure angles from the perpendicular line. Let the angles subtended by the ends of the wire at P with the perpendicular line from P to the wire be $\theta_1$ and $\theta_2$. If the point P is at (0, r) and the wire is from $x_1$ to $x_2$, then $\tan \theta_1 = |x_1|/r$ and $\tan \theta_2 = |x_2|/r$. The integration limits for $\phi$ correspond to the angles the lines from P to the ends of the wire make with the y-axis. Let the angles made by the lines from P to the ends of the wire with the perpendicular be $\alpha$ and $\beta$. If the point is opposite to the center of the wire (from -L/2 to L/2), then $x_1 = -L/2$, $x_2 = L/2$. $E_x = G \lambda \left( \frac{1}{\sqrt{(-L/2)^2 + r^2}} - \frac{1}{\sqrt{(L/2)^2 + r^2}} \right) = G \lambda \left( \frac{1}{\sqrt{L^2/4 + r^2}} - \frac{1}{\sqrt{L^2/4 + r^2}} \right) = 0$. $E_y = \frac{G \lambda}{r} (\sin \phi_2 - \sin \phi_1)$. Here $\tan \phi_1 = (-L/2)/r$ and $\tan \phi_2 = (L/2)/r$. $\phi_1 = -\arctan(L/2r)$ and $\phi_2 = \arctan(L/2r)$. $E_y = \frac{G \lambda}{r} (\sin(\arctan(L/2r)) - \sin(-\arctan(L/2r))) = \frac{G \lambda}{r} (\sin(\arctan(L/2r)) + \sin(\arctan(L/2r)))$. Let $\alpha = \arctan(L/2r)$. Then $\tan \alpha = L/2r$. Consider a right triangle with opposite side L/2 and adjacent side r. The hypotenuse is $\sqrt{(L/2)^2 + r^2}$. $\sin \alpha = \frac{L/2}{\sqrt{(L/2)^2 + r^2}}$. $E_y = \frac{G \lambda}{r} (2 \sin \alpha) = \frac{2 G \lambda}{r} \frac{L/2}{\sqrt{L^2/4 + r^2}} = \frac{G \lambda L}{r \sqrt{L^2/4 + r^2}}$. Substitute $\lambda = M/L$: $E_y = \frac{G (M/L) L}{r \sqrt{L^2/4 + r^2}} = \frac{GM}{r \sqrt{L^2/4 + r^2}}$. This is the magnitude of the field, directed towards the wire.

If the point is perpendicular to one end of the wire (from 0 to L), then $x_1 = 0$, $x_2 = L$. $E_x = G \lambda \left( \frac{1}{\sqrt{0^2 + r^2}} - \frac{1}{\sqrt{L^2 + r^2}} \right) = G \lambda \left( \frac{1}{r} - \frac{1}{\sqrt{L^2 + r^2}} \right)$. $E_y = \frac{G \lambda}{r} (\sin \phi_2 - \sin \phi_1)$. Here $\tan \phi_1 = 0/r = 0$, so $\phi_1 = 0$. $\tan \phi_2 = L/r$, so $\phi_2 = \arctan(L/r)$. $E_y = \frac{G \lambda}{r} (\sin(\arctan(L/r)) - \sin(0)) = \frac{G \lambda}{r} \sin(\arctan(L/r))$. Let $\beta = \arctan(L/r)$. Then $\tan \beta = L/r$. Consider a right triangle with opposite side L and adjacent side r. The hypotenuse is $\sqrt{L^2 + r^2}$. $\sin \beta = \frac{L}{\sqrt{L^2 + r^2}}$. $E_y = \frac{G \lambda}{r} \frac{L}{\sqrt{L^2 + r^2}} = \frac{G (M/L)}{r} \frac{L}{\sqrt{L^2 + r^2}} = \frac{GM}{r \sqrt{L^2 + r^2}}$. This component is directed towards the wire.

The question is general ("gravitational field due to finite wire"). The calculation depends on the position of the point relative to the wire. The formulas derived above are for a point perpendicular to the wire.