Question

The center of mass of a cow and the center of mass of a tractor are 208 meters apart. The magnitude of the gravitational force of attraction between these two objects is calculated to be 1.8 × 10-9 newtons.
What would the magnitude of the gravitational force of attraction be between these two objects if they were 416 meters apart?
A.
1.62 × 10-9 newtons
B.
3.6 × 10-10 newtons
C.
9 × 10-10 newtons
D.
4.5 × 10-10 newtons

Answer 1

Answer:

$4.5\times 10^{-10}\ N$

Explanation:

The gravitational force between two masses is inversely proportional to the square of the distance between them such that,

$\dfrac{F_1}{F_2}=\dfrac{r_2^2}{r_1^2}$ .....(1)

We have,

r₁ = 208 m, $F_1=1.8\times 10^{-9}\ N$, r₂ = 416 m, F₂ = ?

Put all the values in relation (1) such that,

$F_2=\dfrac{F_1r_1^2}{r_2^2}\\\\F_2=\dfrac{1.8\times 10^{-9}\times 208^2}{416^2}\\\\F_2=4.5\times 10^{-10}\ N$

So, the required force is $4.5\times 10^{-10}\ N$. Hence, the correct option is (d).