Question

The planet Mars has a mass of 6.1 × 1023 kg and radius of 3.4 × 106 m. What is the acceleration of an object in free fall near the surface of Mars? The value of the gravitational constant is 6.67259 × 10−11 N · m2 /kg2 . Answer in units of m/s 2 .

Answer 1

The acceleration due to gravity of Mars is $3.5\ m / s^{2}$

Explanation:

As per universal law of gravity, the gravitational force is directly proportional to the product of masses and inversely proportional to the square of the distance between them. But in the present case, the gravity need to be determined between Mars and the object on Mars. Since the mass of Mars is greater than the mass of any object. Thus,

      $\text {Gravitational force of planet}=\frac{G M m}{R^{2}}$

Here, G is the gravitational constant, R is the radius of Mars and M, m is the mass of Mars and the object respectively..

Also, according to Newton’s second law of motion, the acceleration of any object will be equal to the ratio of force exerted on it to the mass of the object.

So in order to determine the acceleration due to gravity of Mars, divide the gravitational force of Mars by mass of object on the surface of Mars.

   $Acceleration\ due\ to\ gravity\ of\ mars =\frac{\text {Gravitational force of Mars}}{m \text { of object }}$

  $Acceleration\ due\ to\ gravity\ of\ mars =\frac{G M m}{R^{2} \times m}=\frac{G M}{R^{2}}$

  $\text { Acceleration due to gravity of mars }=\frac{6.67259 \times 10^{-11} \times 6.1 \times 10^{23}}{3.4 \times 3.4 \times 10^{12}}=\frac{40.703 \times 10^{12}}{11.56 \times 10^{12}}$

 $\text { Acceleration }=3.5\ \mathrm{m} / \mathrm{s}^{2}$