Question

Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon distance of 3.9 × 108 m and the earth’s mass. Assume the orbit is a circle.

Answer 1

Answer: 1010.92 m/s

Step-by-step explanation:

According to Newton's law of universal gravitation:

`F=G(Mm)/(r^(2))` (1)

Where:

`F` is the gravitational force between Earth and Moon

`G=6.674(10)^(-11)(m^(3))/(kgs^(2))` is the Gravitational Constant

`M=5.972(10)^(24) kg` is the mass of the Earth

`m=7.349(10)^(22) kg` is the mass of the Moon

`r=3.9(10)^(8) m` is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth a circular orbit, the Earth exercts a centripetal force on the moon, which is equal to
`F`:

`F=m.a_(C)` (2)

Where
`a_(C)` is the centripetal acceleration given by:

`a_(C)=(V^(2))/(r)` (3)

Being
`V` the orbital velocity of the moon

Making (1)=(2):

`m.a_(C)=G(Mm)/(r^(2))` (4)

Simplifying:

`a_(C)=G(M)/(r^(2))` (5)

Making (5)=(3):

`(V^(2))/(r)=G(M)/(r^(2))` (6)

Finding
`V`:

`V=\sqrt{(GM)/(r)}` (7)

`V=\sqrt{((6.674(10)^(-11)(m^(3))/(kgs^(2)))(5.972(10)^(24) kg))/(3.9(10)^(8) m)}` (8)

Finally:

`V=1010.92 m/s`