# Drag force on a sphere [closed]

A sphere of mass $M$ and radius $R$ moves with speed $V$ through a region of space that contains particles of mass $m$ that are at rest. There are $n$ of these particles per unit volume. Assume $m$ is very small compared to $M$, and assume that the particles do not interact with each other. What is the drag force on the sphere?

I found this problem in the section of conservation of energy in David morin classical mechanics book.

To address this question I firstly thought of the mechanism by which an ideal gas imposes a dragging Force proportional to velocity gradient owing to internal collisions between the body and the gas molecules. But in this case the main difference is that the spheres of mass m are at rest initially so that this approach simply wont do.

Then this idea came to my mind that the body,as it propagates through the medium creates spherical waves in that medium just like the pressure waves in ideal gas and if this is the case then the drag force must be proportional with the acoustic pressure at the region where the body lies.As the position of the body is not fixed,say,at position A it created a spherical wave and now the body has reached point B and at point B the drag force should be according to my logic equal to the acoustic pressure due to the wave which it created while it was it at point A. Now I am failing to mathematically model this idea to turn it into an answer to the question. How can i do this?

## closed as off-topic by sammy gerbil, Jon Custer, Yashas, heather, Qmechanic♦Aug 16 '17 at 6:56

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• I would add the fluid dynamics tag. If you are familiar with comp!ex numbers, this is related: en.wikipedia.org/wiki/Joukowsky_transform – user163104 Aug 13 '17 at 15:36
• Think in terms of energy conservation and where and how the energy flows. There is momentum exchange and alternately heat production. Drag occurs in the latter. – docscience Aug 13 '17 at 15:42

## 1 Answer

This problem was solved by Newton. It is still used today by aerodynamicists to give quick and simple solutions to hypersonic flows, where a body travels at many times the speed of sound. Most people find it easier to consider the equivalent situation of a stationary body with air streaming past it. When the flow is hypersonic, the random motion of molecules (which is what is involved in sound waves) can be neglected compared to their mean velocity. The flow can then be regarded as equivalent to the stationary body being bombarded by a parallel stream of identical particles that bounce off it. In turn this is equivalent to the problem proposed.

Newton was unsure whether the bounces would be perfectly elastic (the normal component of velocity is reversed) or perfectly anelastic (the normal component of momentum is destroyed) or somewhere between. Anelastic collision turns out to be more realistic, and gives results that are often (for simple shapes like spheres) within ten percent of experiment or computational fluid dynamics.

The simple calculation is that if an element dA of a surface is inclined at an angle $\delta$ to the flow, it will be impacted by particles in a stream of cross section $\sin{\delta}dA$, having momentum $\rho V \sin{\delta}dA$, where $\rho$ is the density of the flow and $V$ is its velocity. The component of this momentum normal to the surface that arrives per unit time is $\rho V^2\sin^2{\delta}$ per unit area and this can be thought of as the increase $P$ in fluid pressure. The drag force is the integral of this over the frontal area of the body. For a sphere of radius $R$ this is $$\int_0^{\pi/2}\rho V^2\cos^2{\theta}(2\pi R\sin{\theta}(\cos{\theta}d\theta)=(\rho V^2/2)(\pi R^2)$$ where $\theta$ is measured from the axis of symmetry so that $\delta=\pi/2-\theta$. If elastic collisions are assumed, as is probably the case for the Ops question, the answer is twice as big.

It is usual to assume that pressure on parts of the body not exposed to the flow are negligible and that this expression gives the resistance. This flow is very unlike the flows at low speeds to which complex variables can be related. However, by remarkable coincidence, the predicted resistance is in the right ballpark for these flows also

• Why momentum relates to square of the velocity? – user157588 Aug 13 '17 at 17:38
• We are talking about the rate at which momentum arrives at the body – Philip Roe Aug 13 '17 at 20:13
• But I made an edit to clarify – Philip Roe Aug 13 '17 at 21:17