Question

F = 2.0*10^20 N,

G = 6.67*10^-11 Nm^2/kg^2,
m1 = 5.98*10^24kg,
r = 3.8×10^8m.
solve for m2. I just need detailed help on how to solve it

Answer 1

`F=(Gm_1m_2)/(r^2)`

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

`2.0*10^(20)\,\mathrm N=\frac{\left(6.67*10^(-11)\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98*10^(24)\,\mathrm{kg}\right)m_2}{\left(3.8*10^8\,\mathrm m\right)^2}`

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

`(10^(-11)*10^(24))/((10^8)^2)=(10^(24-11))/(10^(16))=10^(-3)`

Meanwhile, the other values on the right reduce to

`(6.67*5.98)/(3.8^2)\approx2.76`

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

`2.0*10^(20)\,\mathrm N=\left(2.76*10^(-3)\,\frac{\mathrm N}{\mathrm{kg}}\right)m_2`

Now we solve for
`m_2`:

`m_2=\frac{2.0*10^(20)\,\mathrm N}{2.76*10^(-3)\,\frac{\mathrm N}{\mathrm{kg}}}\approx0.725*10^(20-(-3))\,\mathrm{kg}`

`m_2=7.25*10^(22)\,\mathrm{kg}`

or, if taking significant digits into account,

`m_2=7.3*10^(22)\,\mathrm{kg}`