a) $5.979 \times 10^{24} \textrm{ kg}$

b) This is the same as the accepted value to four significant figures.

### Solution video

# OpenStax College Physics, Chapter 6, Problem 33 (Problems & Exercises)

### Calculator Screenshots

*m p*by multiplying both sides by

*r p*

*squared over**g*and then switch the sides around. We get mass of the earth is acceleration due to gravity at the Pole times the radius squared divided by*g*. So that's 9.830 meters per second squared, times 6.371 times ten to the six meters squared divided by 6.673 times ten to the minus eleven Newton meters squared per kilogram squared, giving us this value for the mass, 5.97926 times ten to the twenty-four kilograms and to four significant figures, this is the same as the accepted value for the mass of the earth.## Comments

can you remove the symbol for the little mass when not calculating satellites

Hi coleman, removing the little mass is OK whenever you're calculating the acceleration due to gravity for anything, whether or not it's a satellite. Acceleration due to gravity is $a_g = \dfrac{GM_p}{r^2}$, where $M_p$ is the mass of the planet causing the gravity and $r$ is the distance from the center of the planet to wherever the acceleration is occuring. As an exercise, try plugging in the mass of the Earth for $M_p$ and the radius of the Earth for $r$ and see what you get (it should be $9.8 \textrm{ m/s}^2$ for the acceleration due to gravity on the Earth's surface).

All the best,

Shaun