# Answer to Question #11608 in Mechanics | Relativity for parth

Question #11608

A planet moves around the sun in nearly circulas orbit.its period of revolution "T" depends upon 1)radius "r" of the orbit 2)mass "M" OF THE SUN 3) the gravitational constant "G".SHOW DIMENSIONALLY THAT T SQUARE IS DIRETLY PROPOTIONAL TO r cube

Expert's answer

Using

a = GM/R²,

where R is the radius from the center of the planet to the center of a sun, M is the mass of the sun and G is the Gravitational constant and also using the centripetal force equation

a = W²R

where W is the radians / second velocity of the planet. Combining these equations above and cancelling the a (acceleration) gives

GM/R² = W²R ==> GM/R³ = W².

Assuming that W = 1/T we get

GM/R³ = 1/T² ==> T² = R³/GM.

So, T depends upon the radius "R" of the orbit, mass "M" of the sun and the gravitational constant "G" and T² is directly proportional TO R³.

a = GM/R²,

where R is the radius from the center of the planet to the center of a sun, M is the mass of the sun and G is the Gravitational constant and also using the centripetal force equation

a = W²R

where W is the radians / second velocity of the planet. Combining these equations above and cancelling the a (acceleration) gives

GM/R² = W²R ==> GM/R³ = W².

Assuming that W = 1/T we get

GM/R³ = 1/T² ==> T² = R³/GM.

So, T depends upon the radius "R" of the orbit, mass "M" of the sun and the gravitational constant "G" and T² is directly proportional TO R³.

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