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So I found this problem, I know the gravity of earth $g_0$, I found the orbital speed of one satellite using this equation $\frac{GM_em_s}{(R_e+h)^2}=m_sa$

($M_e$ is the mass of earth, $m_s$ is the mass of the satellite, $R_e$ is the radius of earth, $h$ is the height of the satellite)

and by taking $h=0$ I found the gravity of earth $\frac{GM_e}{R_e^2}=g_0$ so I found $GM_e=R_e^2g_0$ and then I found the orbital speed $v=R_e\sqrt{\frac{g_0}{R_e+h}}$ and then I calculated kepler's third law. but the last questions says deduce the mass of the earth $M_e$, How can I do this by not knowing the gravitational constant?

I'm sorry for bad english, and thanks a lot

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How can I do this by not knowing the gravitational constant?

Sarcasm: You could do what Henry Cavendish did, which was to measure how the horizontal deflection rather heavy balls separated horizontally by a small distance and held in place by torsion rods. The concept of the Newtonian gravitational constant did not exist in Cavendish's time, and would not exist until almost a century after Cavendish's paper on Experiments to determine the Density of the Earth was published. At the time, Cavendish envisioned his experiment as a way to "weigh the Earth".

After the fact (about 80 years after the fact!), physicists saw that Cavendish had measured something much more fundamental than the average density of the Earth. We now see that Cavendish had indirectly measured what we now call the Newtonian gravitational constant. G does not come into play in Newton's formulation of his law of gravitation.

The modern version of the Newtonian law of gravitation does use the Newtonian gravitational constant, which frees you from having to redo Cavendish's experiment.

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  • $\begingroup$ thanks a lot I watched a video about Cavendish and now I understand $\endgroup$ – 18.99 Sep 3 at 11:56
  • $\begingroup$ Note that from a precision perspective, the products $GM_\text{Earth}$ and $GM_\text{Sun}$ are known much more precisely than the individual terms $G$, $M_\text{Earth}$, or $M_\text{Sun}$. $\endgroup$ – rob Sep 3 at 13:18
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In short, you can't deduce the mass of the Earth using this method unless you know G, and determining G is the hard part! (Involving heavy metal balls and very sensitive force-meters) Fortunately, you don't have to do this, as its value can be easily found in textbooks and online; it is $6.674\times10^{-11}$

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  • $\begingroup$ thanks a lot, I thought I was missing something $\endgroup$ – 18.99 Sep 3 at 11:55

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