Find the acceleration due to gravity on the surface of Mars. Use a value of 6.42 times 10 to the 23rd kilograms for the mass of Mars and a value of 3.390 times 10 to the sixth meters for its radius.
Let’s call the value 6.42 times 10 to the 23rd kilograms 𝑚, and we’ll refer to the radius of 3.390 times 10 to the sixth meters as 𝑟. We want to solve for the acceleration due to gravity on the surface of Mars, which we’ll call 𝑔 sub 𝑚. As we begin working toward a solution, let’s recall two mathematical relationships.
The first is Newton’s second law. This law of motion states that the net force on an object is equal to the object’s mass multiplied by its acceleration. Second, let’s recall Newton’s gravitational law. This law says that the gravitational force between two masses, 𝑚 one and 𝑚 two, is equal to capital 𝐺, the universal gravitational constant, times the product of the two masses divided by the distance between them squared.
If we apply this gravitation law to our scenario, then the gravitational force of attraction between the planet Mars, represented by the mass 𝑚, and a test mass on the surface of the planet, represented by 𝑚 sub test, is equal by Newton’s second law to the mass of the test mass, 𝑚 sub test, times its acceleration, 𝑎.
It’s this acceleration 𝑎 that is the acceleration of the test mass due to the gravitational field created by Mars. In other words, 𝑎 can be replaced by 𝑔 sub 𝑚, the acceleration due to gravity on Mars’s surface.
Looking at this equation, we see that the test mass, 𝑚 sub test, cancels out leaving us with terms we were either given such as 𝑚 and 𝑟 or constant values such as capital 𝐺, the universal gravitational constant, which we treat as exactly 6.67 times 10 to the negative 11th meters cubed per kilograms second squared.
When we plug-in our values for 𝑔 𝑚 and 𝑟 and enter these values in our calculator we find a number for 𝑔 sub 𝑚 of 3.73 meters per second squared that’s the acceleration due to gravity on the surface of Mars.