Question

how did newton use the moton of the moon around the earth to test his universal (centripetal acceleration of the moon around the earth is 2.7 times 10 - 3 m/s 2)

Answer 1

Answer:

hey mate

Explanation:

here is your answer

Newton compared the acceleration of the moon 2.7 × 10 -³ m/s² with the acceleration during free fall near the surface of the earth 9.8 m/s² and concluded that the acceleration due to earth gravitational force must be decreasing with the distance of the body from the earth surface

$\frac{a \: object}{a \: moon} = \frac{9.8 \: {ms}^{2} }{2.7 \: \times \: {10}^{ - 3} {m s}^{2} }$

~3600. (eqn 1)

$\frac{distance \: of \: the \: moon \: from \: the \: earths \: centre \: }{distance \: of \: the \: object \: from \: the \: earth \: centre}$

= r/R

$= \frac{3.85 \: \times \: {10}^{ - 5} km}{6378 \: km} = 60$

(eqn 2)

= a object/a moon = [r/R]². (eqn 3)

a object is proportional to 1/R². (eqn 3)

from Newton second law of motion F = ma thus the force exerted by the earth on an object of mass m at distance R from it is given as

Fo is proportional to M /R²

In the same way the object also exerts a force on the earth of mass M

Fe is proportional to M/R²

by Newton third law of motion the force exerted by the earth on an object and the force exerted by the object on the earth are equal

thus the force between the earth and the object is F is proportional to Mm/R²

This force was then generalized to gravitational force between any two object

$thanku$