Question

What is the magnitude of the gravitational force on Mars, with a mass of $$6.34\times {10}^{23}$$and a radius of $$3.43\times {10}^{6}m$$?

Answer 1

Hint: The gravitational force between two bodies form an action and reaction pair, that is, the forces are equal in magnitude but opposite in direction. The gravitational force equals the gravitational constant times the product of masses by the square of the distance between them.

Formula Used
$$F={\displaystyle \frac{GMm}{r^2}}$$

Complete step-by-step solution:
From the given information, we have the data as follows.
The mass of the planet Mars is, $$6.34\times {10}^{23}$$and the radius of the planet Mars is, $$3.43\times {10}^{6}m$$.
The force of attraction between any two bodies in the universe is known as the force of gravitation. The force of gravitational attraction between the two bodies acts along the line joining their centre. This force is mutual.
The formula to be used to solve this problem is given as follows.
$$F={\displaystyle \frac{GMm}{r^2}}$$
Where G is the gravitational constant, M and m are the masses and r is the distance between the masses.
As we are supposed to find the magnitude of the gravitational force on the planet Mars, so, we need to edit the above formula as follows.
$${\displaystyle \frac{F}{m}}={\displaystyle \frac{GM}{r^2}}$$
Now we have obtained the formula using which we can compute the value of the magnitude of the gravitational force.
Substitute the values of the mass and the radius of the planet Mars in the above equation.
$${\displaystyle \frac{F}{m}}={\displaystyle \frac{6.673\times {10}^{-11}\times 6.34\times {10}^{23}}{{(3.43\times {10}^6)}^2}}$$
Therefore, the value of the magnitude of the gravitational force on Mars is $$3.597Nk{g}^{-1}$$
$$\therefore$$The value of the magnitude of the gravitational force on Mars is$$3.597Nk{g}^{-1}$$.

Note: The gravity is different from the gravitation. Gravitation is the force attraction acting between any two bodies of the universe, whereas, gravity is the earth’s gravitational pull on a body lying on the near surface of the earth. As the units of the parameters are in SI, so, no need to change them.