Question

Planet X has a mass of M and a radius of R. Planet Y has a mass of 3M and a radius of 3R. Identical satellites orbit both planets at a distance R above their surfaces, as shown above. The planets are separated by such a large distance that the gravitational forces between them are negligible.

How does the magnitude of the gravitational force FY exerted by Planet Y on its satellite compare to the gravitational force FX exerted by Planet X on its satellite?

Answer 1

Answer:

${FY} = \dfrac{3}{4} \times FX$

Explanation:

The parameters given for the planets are;

The mass of planet X = M and the radius of planet X = R

The mass of planet Y = 3·M and the radius of planet Y = 3·R

The magnitude of the gravitational force of the planets on their satellites are given by the following equation;

$F=G \times \dfrac{M_{1} \cdot m_{2}}{R^{2}}$

Where;

M₁ = The mass of the first object = The mass of the planet

m₂ = The mass of the second object = The mass of the satellite

R = The distance between the centers of the two planets = The distance between the center of the planet and the satellite

G = The universal gravitational constant

The force between planet X and the satellite in its orbit = $FX=G \times \dfrac{M \times m}{(2 \cdot R)^{2}} = G \times \dfrac{M \times m}{4 \cdot R^{2}}$

The force between planet Y and the satellite in its orbit = $FY=G \times \dfrac{3\cdot M \times m}{(4 \cdot R)^{2}} = G \times \dfrac{3\cdot M \times m}{16 \cdot R^{2}} = G \times \dfrac{ 3\cdot M \cdot m}{16 \cdot R^{2}}$

Therefore;

$\dfrac{FY}{FX} = \dfrac{G \times \dfrac{ 3\cdot M \cdot m}{16 \cdot R^{2}}}{G \times \dfrac{M \times m}{4 \cdot R^{2}}} = \dfrac{3}{16} \times \dfrac{4}{1} = \dfrac{3}{4}$

${FY} = \dfrac{3}{4} \times FX$