Gravitational field intensity is the force experienced by 1 Kg mass in the gravitational field of any massive body. Mathematically, the magnitude of \(E_g=\frac{GM}{r^2}\) Since gravitational Force acts toward the center of the earth

Gravitational Field Intensity (E_g) = -GM/r² \(\widehat r\)

Solution:

Let, radius of earth be Re = 63.8 x 10^5 m

Let, the gravitational field intensity of earth before squeezed be Ei

Now according to the formula, \(E_i=\frac{GM}{R_e^2}\) 

Or, \(E_i=\frac{GM}{{(63.8\times10^5)}^2}\) ---------- (1)

Now gravitational field intensity of earth after being squeezed to the size of moon is

\(E_f=\frac{GM}{R_m^2}\)     [ Rm is the radius of moon = 1.7 x 10^6 m]

Or, \(E_f=\frac{GM}{{(1.7\times10^6)}^2}\) --------- (2)

Now dividing equation 2 by equation 1 we get,

\(\frac{E_f}{E_i}=\frac{\displaystyle\frac{GM}{R_m^2}}{\displaystyle\frac{GM}{R_e^2}}\)

Or, \(\frac{E_f}{E_i}=\frac{R_e^2}{R_m^2}\)

Or, \(\frac{E_f}{E_i}=\frac{{(63.8\times10^5)}^2}{{(1.7\times10^6)}^2}\)

Or, \(\frac{E_f}{E_i}=14\)

∴ \(E_f=14E_i\)

Hence the gravitational field intensity of the earth surface will increase by around 14 times, if earth were to squeezed to the size of the moon.