# Question a38ce

Oct 26, 2016

$r = \sqrt{\frac{G {m}_{1} {m}_{2}}{F}}$

#### Explanation:

To solve for $r$ means you need to make $r$ the subject of the formula.

The problem is that it is the denominator!
There are two ways to fix this.

There is only one term on each side so we can do the following:

$1.$ Cross multiply to give :

${r}^{2} \times F = G {m}_{1} {m}_{2} \text{ } \leftarrow \div$ by F

${r}^{2} = \frac{G {m}_{1} {m}_{2}}{F} \text{ } \leftarrow$ now find the square root.

$r = \sqrt{\frac{G {m}_{1} {m}_{2}}{F}} \text{ }$ all values are positive

$2.$ Invert the whole formula (1 term = 1 term, so it's OK)

$\frac{1}{F} = {r}^{2} / \left(G {m}_{1} {m}_{2}\right) \text{ } \leftarrow$ isolate ${r}^{2} \text{ by } \times G {m}_{1} {m}_{2}$

$\frac{G {m}_{1} {m}_{2}}{F} = {r}^{2} \text{ } \leftarrow$ now find the square root

$r = \sqrt{\frac{G {m}_{1} {m}_{2}}{F}} \text{ }$ all values are positive

Oct 26, 2016

See explanation.

#### Explanation:

Rearranging,

r_(12)=sqrt((Gm_1m_2)/F, where

the universal 4-sd gravitational constant

$G = 6.674 X {10}^{- 20} {\left(k m\right)}^{3} / \left(\left(k g\right) {s}^{2}\right)$,

${m}_{1} \mathmr{and} {m}_{2}$ are masses and ${r}_{12}$ is the distance between

the centers of the masses of which one (say ${m}_{2}$ ) should attract

the bodies around.

If ${m}_{2}$ attracts ${m}_{1}$, then,

$F = {m}_{1}$ X (acceleration g, due to gravity of ${m}_{2}$)

Elucidation:

I have used data from

https://www.newscientist.com/article/dn24068-gravity-map-reveals-earths-extremes/

$g = 9.764$ meter/${s}^{2}$=0.009764 km/${s}^{2}$ = ${m}_{1} / F$

The mass ${m}_{2}$ of the attracting Earth = $5.972 X {10}^{24}$kg.

Here, r_(12) = the radius from the center of the Earth to the point

under reference.

So, the radius of the Earth here

= sqrt( ((6.674 X 10^(-20) X (5.972 X 10^(24)))/0.009764) km

=6389 km.

Similar approximations for Arctic surface gives the radius

to Arctic as

sqrt( ((6.674 X 10^(-20) X (5.972 X 10^(24)))/0.009834) km

= 6366 km.