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Along the x-axis, at distance x from the origin, the gravitational field due to a mass distribution is A*x/(x² + a²)^(3/2) directed along x. What is the magnitude of the corresponding gravitational potential at that point (taken zero at infinity)?

1. A/(x² + a²)^(1/2)
2. A/(x² + a²)^(3/2)
3. A*(x² + a²)^(3/2)
4. A*(x² + a²)^(1/2)

Correct answer: A/(x² + a²)^(1/2)

Solution

Gravitational field E = -dV/dx, so V = -integral E dx. With E = A*x/(x²+a²)^(3/2), let u = x² + a², du = 2x dx. Then integral x/(x²+a²)^(3/2) dx = (1/2) integral u^(-3/2) du = (1/2)*(-2)*u^(-1/2) = -1/(x²+a²)^(1/2). So V = -A*(-1/(x²+a²)^(1/2)) + C = A/(x²+a²)^(1/2) + C; with V(infinity)=0, C=0. Magnitude = A/(x² + a²)^(1/2).

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