Gravitational Fields: Question 6

Worked solution

Step 1: Recall Newton’s law of gravitation

For two point masses (or bodies small enough, or far enough apart, to be treated as point masses): $F = \frac{Gm_1m_2}{r^2}$

Step 2: Substitute the values

Here $m_1 = 5.00\times10^{13}\text{ kg}$, $m_2 = 2.00\times10^{10}\text{ kg}$, and $r = 4.00\times10^4\text{ m}$: $F = \frac{(6.67\times10^{-11})(5.00\times10^{13})(2.00\times10^{10})}{(4.00\times10^4)^2}$

Working the numerator in stages: $6.67\times5.00=33.35$, and $10^{-11}\times10^{13}=10^2$, so $Gm_1 = 33.35\times10^2=3.335\times10^3$.

Then $3.335\times10^3\times2.00\times10^{10} = 6.67\times10^{13}$ (since $3.335\times2.00=6.67$ and $10^3\times10^{10}=10^{13}$).

Working the denominator: $(4.00\times10^4)^2 = 16.0\times10^8 = 1.60\times10^9$.

$F = \frac{6.67\times10^{13}}{1.60\times10^9} = 4.169\times10^4\text{ N} \approx 4.17\times10^4\text{ N}$

Check by recomputing differently: $6.67/1.60=4.169$ and $10^{13}/10^9=10^4$, so $F=4.169\times10^4\text{ N}$, matching exactly.

Why the other options are wrong

• A ($1.67\times10^9\text{ N}$): this comes from forgetting to square the separation, i.e. calculating $Gm_1m_2/r$ instead of $Gm_1m_2/r^2$, which gives a value with a far too large order of magnitude.
• B ($4.17\times10^3\text{ N}$): this is the correct value divided by $10$, a power-of-ten slip when combining the exponents.
• D ($1.04\text{ N}$): this comes from mistakenly cubing the separation instead of squaring it, i.e. calculating $Gm_1m_2/r^3$, which gives a value that is far too small.

Final answer

• The gravitational force of attraction between P and Q is $F=\boxed{4.17\times10^4}\text{ N}$, option C.