B Is F = -mg a mistake in orbital mechanics?

[Philip Robotic]

Hi everyone!

I'm really sorry if I'm using the wrong forum. It's my first time at PF.

I'm pretty new to physics, as I began studying it just two years ago, but I'm really interested in the subject of astrophysics so I bought myself and started reading 'Introduction to rocket science and engineering' by Travis S. Taylor. So far I'm really enjoying it, but at the beginning of orbital mechanics, where the author also shows the basic dynamic equations related to gravity.

And the third one looks weird. It says: F = -mg

And as far as I know, it should look like this: F = mg

Is this a mistake in my book, or there's a specific case in which the F = -mg can be used?

Ps. If I'm using the wrong forum, tell me and I'll copy and delete the thread here and post it on the right one.

 

[anorlunda]

Gravity makes massive objects attract each other. The planet attracts the satellite. The force pulls the satellite toward the planet. But the sign about whether a force in that direction is plus or minus is an arbitrary choice.

Does that answer your question?

 

[Dale]

The correct expression is $\Sigma f=ma$. If they are assuming there is a single external force then $\Sigma f=f$ and if they are further assuming that $a=-g$ then you would get $f=-mg$

As @anorlunda says, look at the direction of the force and at the direction of the coordinate system to see if the $a=-g$ assumption makes sense.

 

[jtbell]

Different textbooks can use different coordinate systems, sign conventions and systems of units, which can affect the form of equations. You have to be aware of the context that the author is working in.

Another detail that students often overlook is whether a quantity in an equation is a scalar (usually italic like $g$) or a vector (sometimes boldface $\mathbf{g}$ or with an arrow on top $\vec g$ or maybe something else), which can affect which sign is appropriate.

 

[Philip Robotic]

Wow, I didn't expect such a quick reply! Thank you very much for explaining!

 

[Chestermiller]

If upward (pointing) force is defined as positive, then F = -mg. If downward (pointing) force is defined as positive, then F = + mg. So it depends on whether F is defined as positive upwards or positive downwards. If you describe the force vectorially using unit vectors, you can never go wrong. $\vec{F}=mg(-\vec{i}_z)$, where $\vec{i}_z$ is the unit vector in the upward (z) direction.