Finding mass of Earth using the Moon and Kepler

[hmvince]

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:

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:

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

 

[Creator]

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:

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:

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 :
m_e+ m_m = (4*(pi)²*r³) / (G*t²)

Creator

 

[hmvince]

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