Mass of object in Escape velocity

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Escape velocity is primarily determined by the mass of the central body and the distance from its center, not the mass of the escaping object. The formula for escape velocity includes both the mass of the central body and the escaping body, but the latter's contribution is negligible for practical scenarios. For example, the mass of an object would need to be extraordinarily large, around 10^15 kg, to significantly affect the escape velocity from Earth. This is equivalent to the mass of about 100 mountains, making it impractical in real-world applications. Thus, the escape velocity remains effectively constant regardless of the mass of the escaping object.
asmalik12
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Why doesn't the escape velocity depend upon the mass of body?
 
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asmalik12 said:
Why doesn't the escape velocity depend upon the mass of body?
Why doesn't the height reached by a tossed ball depend on the mass of the ball? Same thing. (As is often the case when dealing with gravity, the mass drops out.)
 
asmalik12 said:
Why doesn't the escape velocity depend upon the mass of body?
It does. The correct equation is

v_e = \sqrt{\frac{2\,G(M+m)}r}

The capital M denotes the mass of the central body, the little m, the mass of the escaping body.

Now think about how big that little m needs to be to make any measurable difference in the velocity needed to escape Earth orbit. Escape velocity from LEO is about 11 km/s. Suppose we can measure velocity down to the micrometer/second range, or 10 significant digits. That corresponds to a mass of 1015 kg, or the mass of about 100 mountains.

We aren't going to launch a 100 mountain mass out of Earth orbit any time soon.
 

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