Find the acceleration due to gravity on the surface of Mars. The mass of Mars is 6.42x10²³ kg and its radius is 3370 km. Ans. (3.77 ms⁻²)


Given:  

Mass of the Mars = M = 6.42 ｘ 10²³ kg
Radius of the Mars = R = 3370 km = 3.370 ｘ 10³ｘ 10³ m = 3.370 ｘ 10⁶ m
Gravitational Constant = G = 6.67 ｘ 10⁻¹¹ N m² kg⁻²

To Find:
The magnitude of  Gravitational acceleration on the surface of Mars  = g =  ? m s⁻²

Solution:   

Let m be the mass of the body on the surface of Mars and M is the mass of mars. The distance between their centres is equal to the radius R of Mars then According to the Law of Gravitation we have 

F = `\frac {G m M}{R^2 }`   -----eqn (1)

But according to the 2nd law of Newton

F = W = mg ----eqn (2)

By comparing these two Eqns 1 & 2 we have 

mg = `\frac {G m M}{R^2 }`

m will cancel from both sides hence 

g = `\frac {G M}{R^2 }`

putting values

g = `\frac {6.67 ｘ 10⁻¹¹ N m² kg⁻² ｘ 6.42 ｘ 10²³ kg}{(3.370 ｘ 10⁶ m)^2 }`

g = `\frac {6.67 ｘ 10⁻¹¹ ｘ 6.42 ｘ 10²³ }{11.3569 ｘ 10¹²}` m s⁻²

g = `\frac {6.67 ｘ 6.42}{11.3569}`ｘ 10⁻¹¹⁺²³⁻¹² m s⁻²

by simplifying we get

g = 3.77ｘ 10⁰ m s⁻²

or

g = 3.77 m s⁻²

Thus, the gravitational acceleration g on the surface of Mars is 3.77 ms⁻²

************************************

************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2020-21 Academic Skills and Knowledge (ASK)     

Note:  Write me in the comments box below for any query and also Share this information with your class-fellows and friends.