Find the gravitational potential at large distances of a thin circular loop of radius \(a\) and mass \(m\), up to terms of order \(r^{-3}\). Find also the potential, and the leading term in the gravitational field near the origin, at distances \(r \ll a\).

First, let's calculate the mass per unit length of the loop. Since the mass of the entire loop is given as \(m\), mass per unit length, \(\lambda\) will be: \(\lambda = \frac{m}{2\pi a}\)

The gravitational potential at a point due to a mass element is given by: \(V_{element} = \frac{-G \cdot dm}{r}\) where \(G\) is the gravitational constant, \(dm\) is a small mass element, and \(r\) is the distance of the mass element to the point. To find the potential at any point \((x, y, z)\) in the space, we need to integrate the potential contributed by each mass element on the loop: \(V(x, y, z) = -G \int\limits_{\text{loop}} \frac{\lambda dS}{r}\) where \(dS\) is the length of an infinitesimal section of the loop.

At large distances, \((r >> a)\), we can use the approximation: \(r \approx \sqrt{x^2 + y^2 + z^2}\) Now, let's choose a coordinate system with the loop in the \(xy\)-plane, centered at the origin. Given that \(z\) is constant, the formula for the gravitational potential can be written as: \(V(x, y, z) = -G \int\limits_{0}^{2\pi} \frac{\lambda a d\theta}{\sqrt{x^2 + y^2 + z^2 - 2ax\cos\theta - 2ay\sin\theta + a^2}}\) To solve the integral, we will use power series expansion: \(\frac{1}{r} \approx \frac{1}{\sqrt{x^2 + y^2 + z^2}} + \frac{ax\cos\theta + ay\sin\theta}{(x^2 + y^2 + z^2)^{3/2}} - \frac{a^2 (x\cos\theta + y\sin\theta)^2}{2(r^2)^{3/2}}\) Now, substitute the expression for \(r\) and integrate term by term. The first and second terms will integrate to zero, and we are left with only the third term: \(V(x, y, z) = -\frac{Gm}{2} \frac{a^2}{(x^2 + y^2 + z^2)^{3/2}}\)

Near the origin, we can ignore terms higher than first order in \(r\), so we can approximate the potential as: \(V(x, y, z) = -G \int\limits_{0}^{2\pi} \frac{\lambda a d\theta}{\sqrt{a^2 - 2ax\cos\theta - 2ay\sin\theta}}\) To solve the integral, we can use the fact that the potential is rotationally symmetric about the \(z\)-axis: \(V(x, y, z) = -G \int\limits_{0}^{2\pi} \frac{\lambda a d\theta}{\sqrt{a^2 - 2ar\cos\theta}}\) Now, replacing \(x \mbox { and } y\) with \(r\cos\phi \mbox { and } r\sin\phi\), \(V(x, y, z) = -G \int\limits_{0}^{2\pi} \frac{\lambda a d\theta}{\sqrt{a^2 - 2ar\cos(\theta - \phi)}}\) Using the approximation \(V(r) \approx V(a)\) for \(r \ll a\), we have: \(V(r) \approx -\frac{GM}{a}\) where \(M\) is the total mass of the loop. The leading term in the gravitational field near the origin can be found using the gradient of the potential, which is proportional to the radial component. Therefore, the field strength will be proportional to the \(r^{-2}\) term, giving: \(g_{leading}(r) = \frac{GM}{a^2} r\) These are the expressions for the gravitational potential and leading term in the gravitational field near the origin and at large distances of a thin circular loop with radius \(a\) and mass \(m\).