Question

. Neptune is 1.63 × 1012 meters from Uranus.Calculate the magnitude of the interplanetary force of attraction between Uranus and Neptune at this point. [Show all work, including the equation and substitution with units.]

Answer 1

Given,

The distance between the Neptune and Uranus, r=1.63×10¹² m

From Newton's law of gravitation, the gravitational force between two objects is proportional to the product of the mass of two objects and inversely proportional to the square of the distance between them.

Thus the force of attraction between Neptune and Uranus is given by,

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

Where:

• G=6.67×10⁻¹¹ m³kg⁻¹s⁻² is the gravitational constant.

,

• m₁=1.024×10²⁶ kg is the mass of Neptune.

,

• m₂=8.68×10²⁵ kg is the mass of Uranus.

On substituting the known values,

`\begin{gathered} F=(6.67*10^(-11)*1.024*10^(26)*8.68*10^(25))/((1.63*10^(12))^2) \\ =2.23*10^(17)\text{ N} \end{gathered}`

Thus the interplanetary force of attraction between Neptune and Uranus is 2.23×10¹⁷ N