Need help with reversal of a calculation involving gravitational constant.

[Edwin1974]

Hi.

I would like to know if it is possible to "reverse engineer" the formula below to find (G) if Ve, M and r are all known values.

Ve = sqrt{2GM/r}

Ve would be escape velocity, G would be gravitational constant, M would be mass of planet and r would be radius of planet.

I hope I am explaining myself correctly here.

In other words how would I go about reversing the formula Ve = sqrt(2GM/r) so that is begins with G =

Kind Regards
Edwin.

 

[MarkFL]

Hello, and welcome to MHB, Edwin! (Wave)

We are given:

$$V_e=\sqrt{\frac{2GM}{r}}$$

What do we get if we square both sides of the equation?

 

[Edwin1974]

Hi Mark.

Thank you for replying to my question. If I square the one side of the equation I get 125075572.2797 and I arrive at the same answer if I square the other side.

I should just let you know that my math skill is equal to about 10 to the power of minus 11 ok, so really struggling with this one. I am not really sure what you are trying to show me with squaring both sides of the equation, I see that by doing that I arrive at the same answer but don't know how that knowledge is used to reverse calculate Ve = sqrt{2GM/r} so that the equation can begin with G = . I hope I am making sense, if not please let me know.

Kind Regards
Edwin.

 

[DrWahoo]

After you square both sides, make sure to multiply by $r$ to get rid of the fraction.

Once you do that, you should have a simple equation to solve using multiplication and division.

What are you getting as a solution without plugging in numbers for each unknown?

 

[Greg]

$$V_e=\sqrt{\frac{2GM}{r}}$$

$$V_e=\left(\frac{2GM}{r}\right)^{1/2}$$

$$V^2_e=\left(\frac{2GM}{r}\right)^{1/2}\cdot\left(\frac{2GM}{r}\right)^{1/2}=\left(\frac{2GM}{r}\right)^{1/2+1/2}=\left(\frac{2GM}{r}\right)^1=\frac{2GM}{r}$$

$$V^2_e\cdot\frac{r}{2M}=\frac{2GM}{r}\cdot\frac{r}{2M}$$

Can you continue?

 

[Edwin1974]

Thank you for all your help here guys.

G = (Ve^{2})/2M

Kind Regards
Edwin

 

[MarkFL]

Thank you for all your help here guys.

G = (Ve^{2})/2M

Kind Regards
Edwin

It's actually:

$$G=\frac{rV_e^2}{2M}$$