Question 6.6: (a) Find the acceleration due to Earth’s gravity at the dist...

Solution for (a)

Substituting known values into the expression for g found above, remembering that M is the mass of Earth not the Moon, yields

g = G\frac{M}{r^2} =\left( 6.67×10^{−11} \frac{N ⋅ m^2}{kg^2}\right) × \frac{5.98×10^{24} kg}{(3.84×10^8 m)^2}                  (6.46)

Solution for (b)

Given that the period (the time it takes to make one complete rotation) of the Moon’s orbit is 27.3 days, (d) and using

1 d×24\frac{hr}{d} × 60\frac{min}{hr} × 60 \frac{ s}{min} = 86,400 s                   (6.49)

we see that

ω = \frac{Δθ }{Δt} = \frac{2π rad }{(27.3 d)(86,400 s/d)} = 2.66×10^{−6}\frac{rad}{s} .                  (6.50)

The centripetal acceleration is

a_c = rω^2 = (3.84×10^8 m)(2.66×10^{−6} rad/s)^2                        (6.51)

The direction of the acceleration is toward the center of the Earth.
Discussion
The centripetal acceleration of the Moon found in (b) differs by less than 1% from the acceleration due to Earth’s gravity found in (a). This agreement is approximate because the Moon’s orbit is slightly elliptical, and Earth is not stationary (rather the Earth-Moon system rotates about its center of mass, which is located some 1700 km below Earth’s surface). The clear implication is that Earth’s gravitational force causes the Moon to orbit Earth.