How does change in aircraft's weight equal fuel mass flow rate multiplied by 'g'?

Can someone help me understand how does $$\frac{dW}{dt} = -\dot{m}_f \cdot g ,$$ where $$\dot{m}_f$$ is fuel mass flow rate. This is the first equation laid down in the Breguet range equation derivation.

Here's where I encountered it. Alternative accessible source (link):

In the SI system the weight is measured in Newtons [N = kg*m/s^2], while the mass is measured in kilograms [kg].

To convert kilograms in Newtons, i.e. a mass measurement in a weight, you multiply by the acceleration of gravity g.

In this case you're converting a change in mass to a change in weight, the conversion is the same, but both sides have an additional time component in the denominator due to the derivative.

Given your comment, it seems you're confused by the good old mass-vs-weight thing. Let's have Wikipedia:

In common usage, the mass of an object is often referred to as its weight, though these are in fact different concepts and quantities. In scientific contexts, mass is the amount of "matter" in an object (though "matter" may be difficult to define), whereas weight is the force exerted on an object by gravity. In other words, an object with a mass of 1.0 kilogram weighs approximately 9.81 newtons on the surface of the Earth, which is its mass multiplied by the gravitational field strength. The object's weight is less on Mars, where gravity is weaker, and more on Saturn, and very small in space when far from any significant source of gravity, but it always has the same mass.

• You have explained the dimensionality of the components, all right. But I am trying to understand how does mass flow rate, which from my thermodynamics class equates $\rho Av$, multiplied by gravity equates to $dW/dt$! Unless $m_f$ means something completely different here. Commented Nov 12, 2020 at 15:53
• @hegemon8 yeah, mass flow rate can also be represented as $\rho A v$, but it is still $\dot m$. And the relationship between $\dot m$ and $\dot W$ is a simple factor of $g$. Mass is one thing, Weight is a measure of the force felt by the mass in a gravitational field. Commented Nov 12, 2020 at 16:11
• @hegemon8, density $\rho$ is mass per volume ([kg/m³]), $A$ is area [m²] and velocity is distance per time ([m/s]), so multiplying them together you get mass per time ([kg/s]), the mass flow rate. Commented Nov 13, 2020 at 21:59
• @Federico Oh, yes. Quite silly of me actually. I was extrapolating things where none's necessary. Thanks. Commented Nov 14, 2020 at 3:53

The equation you've posted is saying that planes get lighter because they burn fuel. When you mathematically model aircraft, that's often the only significant weight difference, and understanding what the aircraft weighs is important for knowing how much lift the aircraft needs to produce, which changes the drag, and so on...

Treat that equation as a statement that FOR THAT MATHEMATICAL MODEL the ONLY change in the aircraft's weight is caused by it using fuel.

If you were modeling combat aircraft that will be expending weapons, or transport aircraft that air-drop cargo (like a skydiving aircraft), that equation might change to include terms for those payload weights as well.

• Sincerely I don't see how this addresses OP's question Commented Nov 12, 2020 at 20:27
• @Federico you've already seen the comment on your answer. It's clear the OP doesn't understand why the weight change equals the fuel mass flow. This is why. Commented Nov 13, 2020 at 0:03
• yes, they're confused by the transformation of a delta mass into a delta weight. I don't see how it helps them (or how it provides an explanation) talking about payloads and so on. But I gues we'll see if/when they comment Commented Nov 13, 2020 at 10:04
• I talk about the payloads because a student taking a course including Breguet Range equations is likely to be asked about aircraft range in situations that involve more than just fuel. That's the way it worked in my aerospace undergrad. Commented Nov 13, 2020 at 21:12