Question

Is centre of gravity of her head at the same point as centre of mass?

• A. Yes
• B. No
• C. Cannot determine
• D. none of the above

Answer 1

The correct option is B No

We have,
$r_1=120-6\times {10}^{-5}$
$r_2=120+6\times {10}^{-5}$
for center of mass,
$r=\frac{m_1{r}_1+m_2{r}_2}{m_1+m_2}$
$r=\frac{r_1+r_2}{2}$
$=\frac{120-6\times {10}^{-5}+120+6\times {10}^{-5}}{2}$
Now,
for center of gravity,
$r=\frac{F_1{r}_1+F_2{r}_2}{F_1+F_2}$
where,
$F_1=\frac{GmM}{r_1^2}$
$\therefore r=\frac{\frac{1}{r_1}+\frac{1}{2_2}}{\frac{1}{r_1^2}+\frac{1}{r_2^2}}$
$=\frac{(r_2+r_1)r_1{r}_2}{r_1^2+r_2{r}^2}$
$=\frac{\left(240\right)({120}^2-36\times {10}^{10})}{2({120}^2+36\times {10}^{-10})}$
$r\left(\frac{{120}^2-36\times {10}^{-10}}{{120}^2+36\times {10}^{10}}\right)$
i.e. Less than r i.e. less than 120 km