# Earth Gravity vs Mars Gravity

At what altitude/distance/orbit of Earth would the force of gravity equate that of what Mars' gravity is calculated to be?

Furthermore, is the resulting distance within our normal field(s) of orbit used by modern space craft?

It's for a research paper, so sources would be helpful. I can't find any - that's why I'm asking here.

• Well, how do you calculate the force of gravity? How can you set it up, to be the same strength as that of Mars? braeunig.us/space/constant.htm – DilithiumMatrix Nov 27 '16 at 6:07
• @DilithiumMatrix: I appreciate your trying to encourage my own discovery, but I don't seem to be able to wrap my head around how to do the calculation. – Graham Lewis Nov 27 '16 at 6:25
• Then that's the question you should be asking. – DilithiumMatrix Nov 27 '16 at 17:08

If $M_E$ is the mass of the planet Earth, $M_M$ the mass of planet Mars and $r_M$ it's radius, then you have to solve for the distance from the center of Earth's mass $r_E$ where you would have the same gravity as on Mars's surface:

$$\frac{G \cdot M_E}{r_E^2}=\frac{G\cdot M_M}{r_M^2}$$

so $G$ cancels out and you get

$$r_E=\sqrt{\frac{M_E}{M_M}}\cdot r_M$$

which is in meters

$$r_E = 10^7 \text{m} \to h \approx 4\cdot 10^6 \text{m}$$

To feel the force of gravity you nevertheless can't simply take an orbit at this distance, since if you are in orbit you are in free fall and don't feel any forces except tidal forces which we don't need here, so the calculation is assuming stationary observers.

But you could build a tower of height $h$. If you build it on one of the poles you won't have centrifugal counterforces due Earth's rotation, while on the equator you can take them into account to save a few meters, and the equation becomes

$$\frac{G \cdot M_E}{r_E^2} - \omega_E^2\cdot r_E=\frac{G\cdot M_M}{r_M^2}$$

where $\omega_E$ is the angular velocity of planet Earth's spin. The higher the tower, the higher the effect of Earth's rotation in form of gravity-compensating centrifugal-force. In our example the difference is rather neglible and you can use the first equation.

• So, the effective answer for my first question is about 4⋅10^6 m. Also, I'm assuming "m" is in miles. Are these understandings true; if so, are you able to answer the second question: "is the resulting distance within our normal field(s) of orbit used by modern space craft?" – Graham Lewis Nov 27 '16 at 13:09
• The m is for meter (in physics you always use SI-units, not imperial ones), and h=4e6 is the height of the tower as measured from the surface of the earth to the top of the building. Modern spacecraft can fly much higher, but when they are in orbit the crew is in constant free fall and doesn't feel the gravity. It would have to hover at that height, which costs a lot more fuel than just getting the spacecraft into orbit. – Gendergaga Nov 27 '16 at 14:01