Question

A satellite revolving in a circular equatorial orbit of radius $R=2.0\times {10}^4$km from west to east appears over a certain point at the equator every $11.6$h. From these data, calculate the mass of the earth. $(G=6.67\times {10}^{-11}N{m}^2)$.

• A. $16\times {10}^{24}$kg.
• B. $26\times {10}^{24}$kg.
• C. $36\times {10}^{24}$kg.
• D. $6\times {10}^{24}$kg.

Answer 1

Answer: (D)

$6\times {10}^{24}$kg.

We know from the previous problem that a satellite moving west to

east at a distance $R=2-00\times {10}^{4}km$ from the center of the earth will be

revolving round the earth with an angular velocity faster then the

earth's diurnal angular velocity. Let

$\omega =$ angular velocity of the stale

${\omega }_0=\frac{2\pi }{T}=$ angular velocity of the earth. Then

$\omega -{\omega }_0=\frac{2\pi }{t}$

as the relative angular velocity with respect to earth.

Now by Newton's law

$\frac{yM}{R^2}={\omega }^{2}R$

So, $M=\frac{R^3}{t}{(\frac{2\pi }{t}+\frac{2\pi }{T})}^2$

$=\frac{4{\pi }^2{R}^3}{y{T}^2}{(1+\frac{T}{t})}^2$

Substitution gives

$M=6.27\times {10}^{24}kg$