Question

A planet moves around the sun is nearly circular orbit its period of revolution t depends upon

1)radius r of orbit.
2)mass m of the sun.
3) gravitational contact. show dimensionally t^2 proportional to

`{r}^(3)`


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Answer 1

t² ∝ r³

Step-by-step explanation:

Let m be the mass of the sun and m₁ be the mass of the planet and r its distance from the sun. Its gravitational force of attraction equals its centripetal force. So,

Gmm₁/r² = m₁rω² = m₁r(2π/t)² were t is the period of orbit of the planet

Gm/r² = 4π²r/t²

rearranging we ave

t² = (4π²/Gm)r³ since k = 4π²/Gm = constant

t² = kr³ and

t² ∝ r³