Need help with reversal of a calculation involving gravitational constant.

In summary, Edwin is trying to reverse engineer the formula Ve = sqrt(2GM/r) to find the value of G. Mark suggests squaring both sides of the equation and then multiplying by r to get rid of the fraction. Edwin then arrives at the solution G = (Ve^2)/2M.
  • #1
Edwin1974
15
2
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.
 
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  • #2
Hello, and welcome to MHB, Edwin! (Wave)

We are given:

\(\displaystyle V_e=\sqrt{\frac{2GM}{r}}\)

What do we get if we square both sides of the equation?
 
  • #3
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.
 
  • #4
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?
 
  • #5
$$V_e=\sqrt{\frac{2GM}{r}}$$

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

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

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

Can you continue?
 
  • #6
Thank you for all your help here guys.

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

Kind Regards
Edwin
 
  • #7
edwin576 said:
Thank you for all your help here guys.

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

Kind Regards
Edwin

It's actually:

\(\displaystyle G=\frac{rV_e^2}{2M}\)
 

What is the gravitational constant and why is it important in calculations?

The gravitational constant, denoted as G, is a fundamental constant in physics that represents the strength of the gravitational force between two objects. It is important in calculations because it allows us to determine the magnitude of the gravitational force between any two objects in the universe.

What is a calculation involving the gravitational constant and how is it typically used?

A calculation involving the gravitational constant is typically used to determine the force of gravity between two objects. The formula for this calculation is F = (G * m1 * m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

Why would someone need help with a reversal of a calculation involving the gravitational constant?

The need for help with a reversal of a calculation involving the gravitational constant may arise when trying to solve for one of the variables in the formula, such as the distance or mass, given the force of gravity. This can be a challenging task and may require assistance in understanding the mathematical steps involved.

What are some common mistakes people make when trying to reverse a calculation involving the gravitational constant?

One common mistake is not using the correct units for the variables in the formula. The units for the gravitational constant are typically in units of N * m^2 / kg^2, so it is important to ensure that all variables are in the correct units to get an accurate result. Another mistake is not properly rearranging the formula to solve for the desired variable.

Are there any resources available for understanding and solving calculations involving the gravitational constant?

Yes, there are many resources available online and in textbooks that provide step-by-step instructions and examples for solving calculations involving the gravitational constant. Additionally, your professor or peers may also be able to provide assistance and clarification on any questions you may have.

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