Suppose the Earth was only half the size it is now (half the mass and half the radius), what would "g" be?

2 answers

Newtons law:

g= G Me/r^2

Now, if one halves the radius, mass is not half.

Mass=density*volume=density 4/3 PI r^3 so if radius goes down by 1/2, then mass goes to 1/8 Me

So your question is really confusing.
Newton's law of universal gravitation:

where force equals the universal constant of gravity multiplied by mass of the earth and mass of an object divided by the raduis squared

F = G((m1*m2)/(r^2))

Force equals the mass of an object multiplied by the acceleration or 'g'

mg = G((m1*m2)/(r^2))

Mass of the object would cancel out on both sides since infact we do not have a second mass

g = G((m1)/(r^2))

The universal constant of gravity as Newton discovered is:

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

The mass of the Earth is:

mass of the earth = 5.98 * 10^24 kg

The radius of the Earth is:

radius of the earth = 6.38 * 10^6 m

The mass then needs to be divided in half:

5.98 * 10^24 kg/2 = 2.94 * 10^24 kg

The radius then needs to be divided in half:

6.38 * 10^6 m/2 = 3.19 * 10^6 m

The equation for the acceleration or 'g' is:

g = 6.67 * 10^-11 m^3/kg*s^2((2.94 * 10^24 kg)/(3.19 * 10^6 m^2))

Leaving 'g' as:

g = 19.27 m/s^2