Weighing the earth : Youare given the following data : g =9.81 ms^(-2).R_(E)= 6.37 xx10^(6) m, the distance to the moon R = 3.84xx10^(8)m and the time period of the moons revolution is 27 .3 days . Obtain the mass of the earth M_E in two different ways.

Weighing the earth : Youare given the following data : g =9.81 ms^(-2)

1.	Weighing the earth : Youare given the following data : g =9.81 ms^(-2).R_(E)= 6.37 xx10^(6) m, the distance to the moon R = 3.84xx10^(8)m and the time period of the moons revolution is 27 .3 days . Obtain the mass of the earth M_E in two different ways.
Answer» <html><body><p></p>Solution :We know that <br/> `M_E=(gR_E^<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)/G=(9.81 xx(6.37xx10^(6))^2)/(6.67 xx10^(-11))=5.97xx10^(24)kg`<br/> The moon is a <a href="https://interviewquestions.tuteehub.com/tag/satellite-1195060" style="font-weight:bold;" target="_blank" title="Click to know more about SATELLITE">SATELLITE</a> of the <a href="https://interviewquestions.tuteehub.com/tag/earth-13129" style="font-weight:bold;" target="_blank" title="Click to know more about EARTH">EARTH</a>. According to Kepler.s Third law <br/> `T^(2) = (<a href="https://interviewquestions.tuteehub.com/tag/4pi-1882352" style="font-weight:bold;" target="_blank" title="Click to know more about 4PI">4PI</a>^2R^3)/(GM_E) M_(E) =(4pi^(2)R^3)/(GT^2)` <br/> `= (4xx3.14 xx3.14 xx(3.84)^3xx10^(24))/(6.67xx10^(-11) xx(27.3xx24xx60xx60)^(2))=6.02xx10^(24)` kg <br/> Both methods yields almost the same answer , the difference between them being less than <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>% .</body></html>