Question

There is a distance of 3.60 * 10 ^ 7 m between the Earth and a satellite. The magnitude of the gravitational force of attraction between the and the Earth is 400 N. What would be the magnitude of the gravitational force of attraction if the distance between them changed to 1.80 * 10 ^ 7 meters?

Answer 1

Given:

The initial distance between the earth and the satellite, R₁=3.60×10⁷ m

The initial force between the earth and the satellite, F₁=400 N

The changed distance between the earth and the satellite R₂=1.80×10⁷ m

To find:

The changed force between the earth and the satellite.

Step-by-step explanation:

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

Let us assume that the mass of the earth is M and the mass of the satellite is m.

Thus,

`\begin{gathered} F_1=(GMm)/(R_1^2) \\ \Rightarrow F_1R_1^2=GMm\text{ }\rightarrow\text{ \lparen i\rparen} \end{gathered}`

And, the changed force is given by,

`\begin{gathered} F_2=(GMm)/(R_2^2) \\ \Rightarrow F_2R_2^2=GMm\rightarrow\text{ \lparen ii\rparen} \end{gathered}`

From equation (i) and equation (ii),

`\begin{gathered} F_1R_1^2=F_2R_2^2 \\ \Rightarrow F_2=(F_1R_1^2)/(R_2^2) \end{gathered}`

On substituting the known values,

`\begin{gathered} F_2=(400*\left(3.60*10^7\right?^2)/(\lparen1.80*10^7)^2) \\ =1600\text{ N} \end{gathered}`

Final answer:

The magnitude of the gravitational force of attraction between the earth and the satellite after the distance between them was changed is 1600 N.