Question

F = 2.0*10^20 N,

Answer 1

$F=\dfrac{Gm_1m_2}{r^2}$

With the given values of $F,G,m_1,r$, we have

$2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}$

Try dealing with the powers of 10 first: On the right, we have

$\dfrac{10^{-11}\times10^{24}}{(10^8)^2}=\dfrac{10^{24-11}}{10^{16}}=10^{-3}$

Meanwhile, the other values on the right reduce to

$\dfrac{6.67\times5.98}{3.8^2}\approx2.76$

Then taking units into account, we end up with the equation

$2.0\times10^{20}\,\mathrm N=\left(2.76\times10^{-3}\,\dfrac{\mathrm N}{\mathrm{kg}}\right)m_2$

Now we solve for $m_2$:

$m_2=\dfrac{2.0\times10^{20}\,\mathrm N}{2.76\times10^{-3}\,\frac{\mathrm N}{\mathrm{kg}}}\approx0.725\times10^{20-(-3)}\,\mathrm{kg}$

$m_2=7.25\times10^{22}\,\mathrm{kg}$

or, if taking significant digits into account,

$m_2=7.3\times10^{22}\,\mathrm{kg}$