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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 Ă— 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 Ă— 10-9 newtons
B.
3.6 Ă— 10-10 newtons
C.
9 Ă— 10-10 newtons
D.
4.5 Ă— 10-10 newtons

Respuesta :

Answer:

Choice: D.

Explanation:

Newton’s Law of Universal Gravitation

It states that objects attract each other with a force that is proportional to their masses and inversely proportional to the square of the distance between them.  The formula is:

[tex]\displaystyle F=G{\frac {m_{1}m_{2}}{r^{2}}}[/tex]

Where:

m1 = mass of object 1

m2 = mass of object 2

r     = distance between the objects' center of masses

G   = gravitational constant: [tex]6.67\cdot 10^{-11}~Nw*m^2/Kg^2[/tex]

The center of mass of a cow is at a distance of r=208 m from the center of the mass of a tractor. The gravitational force of attraction is [tex]F=1.8\cdot 10^{-9}\ N[/tex]. It's required to find the new force of attraction F' if the distance was r'=416 m.

It can be noted that r'=2r. If the distance is doubled, the new force is:

[tex]\displaystyle F'=G{\frac {m_{1}m_{2}}{(2r)^{2}}}[/tex]

[tex]\displaystyle F'=\frac{1}{4}G{\frac {m_{1}m_{2}}{r^{2}}}[/tex]

Substituting the original value of F:

[tex]\displaystyle F'=\frac{1}{4}\cdot 1.8\cdot 10^{-9}\ N[/tex]

Operating:

[tex]\displaystyle F'=4.5\cdot 10^{-10}\ N[/tex]

Choice: D.