Finding mass of Earth using the Moon and Kepler

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SUMMARY

The calculation of Earth's mass using the Moon and Kepler's 3rd Law is accurately represented by the formula me + mm = (4*(pi)²*r³) / (G*t²). In this context, me represents the mass of the Earth, and mm represents the mass of the Moon. The initial calculation yields me = 6.07E+24 kg, but after accounting for the Moon's mass (7.36E+22 kg), the corrected mass of the Earth is 5.99E+24 kg, which aligns closely with the known value of 5.97E+24 kg.

PREREQUISITES
  • Understanding of Kepler's 3rd Law of planetary motion
  • Familiarity with gravitational constant (G)
  • Basic knowledge of orbital mechanics
  • Ability to perform calculations involving scientific notation
NEXT STEPS
  • Study the derivation of Kepler's 3rd Law for non-negligible masses
  • Explore the implications of gravitational interactions in multi-body systems
  • Learn about the gravitational constant (G) and its significance in astrophysics
  • Investigate the methods for measuring celestial body masses using orbital data
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Astronomy students, astrophysicists, and educators interested in celestial mechanics and gravitational calculations will benefit from this discussion.

hmvince
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When calculating the mass of the Earth using the moon as a reference and Kepler's 3rd Law, is it correct to subtract the moon's mass after completing the sum:

Code: 
m[SUB]e[/SUB] = (4*(pi)[SUP]2[/SUP]*r[SUP]3[/SUP]) / (G*t[SUP]2[/SUP])

m[SUB]e[/SUB] = (4*(pi)[SUP]2[/SUP]*385000000[SUP]3[/SUP]) / (G*2358720[SUP]2[/SUP])

m[SUB]e[/SUB] = 6.07[SUB]E[/SUB]+24

Should I be subtracting the moons mass to get:

Code: 
m[SUB]e[/SUB] = 6.07[SUB]E[/SUB]+24 - 7.36[SUB]E[/SUB]+22  =   5.99[SUB]E[/SUB]+24

I know its not a big difference but I would like to be doing this correctly.
Thanks very much
 
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hmvince said:
When calculating the mass of the Earth using the moon as a reference and Kepler's 3rd Law, is it correct to subtract the moon's mass after completing the sum:

Code: 
m[SUB]e[/SUB] = (4*(pi)[SUP]2[/SUP]*r[SUP]3[/SUP]) / (G*t[SUP]2[/SUP])

m[SUB]e[/SUB] = (4*(pi)[SUP]2[/SUP]*385000000[SUP]3[/SUP]) / (G*2358720[SUP]2[/SUP])

m[SUB]e[/SUB] = 6.07[SUB]E[/SUB]+24

Should I be subtracting the moons mass to get:

Code: 
m[SUB]e[/SUB] = 6.07[SUB]E[/SUB]+24 - 7.36[SUB]E[/SUB]+22  =   5.99[SUB]E[/SUB]+24

I know its not a big difference but I would like to be doing this correctly.
Thanks very much

Yes, Newton generalized Kepler's 3rd law for non-negligible mass of the orbiting body .
It is given by :
me+ mm = (4*(pi)2*r3) / (G*t2)

Creator
 
Thankyou very much, 5.99E+24 is much more accurate than 6.07E+24 as the mass of the Earth is in fact, 5.97E+24
Appreciate the reply.
 

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