Question

Calculate the magnitude of the gravitational field of the moon at the location of Earth, in meters per square second.

Answer 1

Answer:

Step-by-step explanation:

We need to find magnitude of acceleration due to gravity on earth surface due to moon

$g_{m}=\frac{GM}{R^2}$

where M=mass of moon

R=distance between moon and earth

G=gravitational constant

$G=6.673\times 10^{-11}\ N-m^2/kg^2$

$M=7.35\times 10^{22}\ kg$

$R=3.84\times 10^8\ m$

$a_m=\frac{6.673\times 10^{-11}\times 7.35\times 10^{22}}{(3.84\times 10^8)^2}$

$a_m=3.33\times 10^{-5}\ m/s^2$

Answer 2

The gravitational field is:

F = 3.3*10^(-5) m/s^2

How to get the gravitational field?

The gravitational field at a distance R of an object of mass M is:

$F = \frac{G*M}{R^2}$

Where G is the gravitational constant.

G = 6.67*10^(-11) m^3/kg*s^2

The mass of the moon is:

M = 7.35*10^22 kg.

The average distance between the moon and the Earth is:

R = 384,400,000 m

Replacing that in the gravitational field equation we get:

$F = \frac{6.67*10^{-11} m^3/kg*s^2*7.35*10^{22} kg}{(384,400,000 m)^2}\\\\F = 3.3*10^{-5} m/s^2$

If you want to learn more about gravity, you can read:

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