# B Change in field strength if two variables are incremented at the same time

#### etotheipi

Given the example $g = \frac{GM}{R^{2}}$, we may compute the change in field strength if the mass is changed by a small amount dM to be$$dg = \frac{G dM}{R^{2}}$$and also if R is changed by dR,$$dg = \frac{-2 GM dR}{R^{3}}$$If, however, both the mass and radius are changed by a small amount at the same time, the source I'm using states that the overall change in field strength is simply the sum:$$dg = \frac{G dM}{R^{2}} - \frac{2 GM dR}{R^{3}}$$I was wondering if anyone could explain why this is a valid step. Thank you!

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#### PeroK

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Given the example $g = \frac{GM}{R^{2}}$, we may compute the change in field strength if the mass is changed by a small amount dM to be$$dg = \frac{G dM}{R^{2}}$$and also if R is changed by dR,$$dg = \frac{-2 GM dR}{R^{3}}$$If, however, both the mass and radius are changed by a small amount at the same time, the source I'm using states that the overall change in field strength is simply the sum:$$dg = \frac{G dM}{R^{2}} - \frac{2 GM dR}{R^{3}}$$I was wondering if anyone could explain why this is a valid step. Thank you!
If you make two small changes, then the overall change is the sum of those changes.

You could look at this a little more rigorously by using a taylor series expansion. The cross terms in $dMdR$ will be small compared to the linear terms in $dM$ and $dR$.

• etotheipi

#### etotheipi

If you make two small changes, then the overall change is the sum of those changes.

You could look at this a little more rigorously by using a taylor series expansion. The cross terms in $dMdR$ will be small compared to the linear terms in $dM$ and $dR$.
I just had a go replacing M and R with $M + dM$ and $R + dR$ and then worked out the resulting change, and ended up obtaining that result you stated with $dMdR$. Thanks!

#### Henryk

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hmm, how do you (physically) change mass? (other than purely mathematically)

#### Ibix

hmm, how do you (physically) change mass? (other than purely mathematically)
Solar wind, or just radiation. I believe that some stars occasionally emit shells if matter fir one reason or another (you'd be better asking in Astronomy and Astrophysics for details). As long as the mass distribution remains spherically symmetric then when a shell of mass passes your radius, its gravity no longer affects you (look up the Shell Theorem).

#### PeroK

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hmm, how do you (physically) change mass? (other than purely mathematically)
You don't necessarily need to change the mass. You might simply want to look at the effect of small variations in mass (to consider experimental error, for example) on the result of your calculations.

• Ibix

#### Henryk

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You don't necessarily need to change the mass. You might simply want to look at the effect of small variations in mass (to consider experimental error, for example) on the result of your calculations.
Ok, that's different.