Design an activity to prove that acceleration due to gravity is not dependent on the mass of the falling object.
Answer
Verified
448.8k+ views
Hint : In this question, use the concept of the gravitational force and equate the force of attraction of the earth with the weight of a falling object. That will prove the required condition. The value of the the Newton's gravitational constant is $ G{\text{ }} = 6.67{\text{ }}x{\text{ }}{10^{ - 11}}{\text{N}}{{\text{m}}^{\text{2}}}{\text{/k}}{{\text{g}}^{\text{2}}} $ .
Complete step by step answer
Let us consider a body of mass $ m $ falling from a height $ h $ on earth’s surface.
Hence the weight of the body is $ mg $
Where, $ g $ is the acceleration due gravity of the earth.
Now we know that, if there are two objects of masses $ M $ and $ m $ respectively placed at a distance of $ r $ then the force acting between them will be-
$ \Rightarrow F = \dfrac{{GMm}}{{{r^2}}} $
Where, $ G $ is the Newton's gravitational constant, $ G{\text{ }} = 6.67{\text{ }}x{\text{ }}{10^{ - 11}}{\text{N}}{{\text{m}}^{\text{2}}}{\text{/k}}{{\text{g}}^{\text{2}}} $
So here in this case,
The force of attraction with which the earth attracts the falling object towards itself is,
$ \Rightarrow F = \dfrac{{GMm}}{{{{\left( {R + h} \right)}^2}}} $
Where, $ R $ is the radius of the earth and h is the distance above the earth’s surface from where the object is being thrown.
But, as we know $ R > > > > h $ , hence the above equation boils down to-
$ \Rightarrow F = \dfrac{{GMm}}{{{{\left( R \right)}^2}}} $
Now equating it with the weight of the object it becomes,
$ \Rightarrow \dfrac{{GMm}}{{{{\left( R \right)}^2}}} = mg $
Solving further we get,
$ \Rightarrow \dfrac{{GM}}{{{R^2}}} = g $
Which is the value of the acceleration due to gravity and from the above equation it is clear that $ g $ is independent from the mass of the falling object.
Note
Please note that the distance between the object’s surface is considered negligible, since the radius of the Earth is very large. If the same concept is applied to a large considerable distance, then the value of g will be a different quantity. Also, we can derive from the above activity that whether a soft feather is being thrown down to earth or a heavy metal ball is thrown, the acceleration due to gravity will be the same in both cases.
Complete step by step answer
Let us consider a body of mass $ m $ falling from a height $ h $ on earth’s surface.
Hence the weight of the body is $ mg $
Where, $ g $ is the acceleration due gravity of the earth.
Now we know that, if there are two objects of masses $ M $ and $ m $ respectively placed at a distance of $ r $ then the force acting between them will be-
$ \Rightarrow F = \dfrac{{GMm}}{{{r^2}}} $
Where, $ G $ is the Newton's gravitational constant, $ G{\text{ }} = 6.67{\text{ }}x{\text{ }}{10^{ - 11}}{\text{N}}{{\text{m}}^{\text{2}}}{\text{/k}}{{\text{g}}^{\text{2}}} $
So here in this case,
The force of attraction with which the earth attracts the falling object towards itself is,
$ \Rightarrow F = \dfrac{{GMm}}{{{{\left( {R + h} \right)}^2}}} $
Where, $ R $ is the radius of the earth and h is the distance above the earth’s surface from where the object is being thrown.
But, as we know $ R > > > > h $ , hence the above equation boils down to-
$ \Rightarrow F = \dfrac{{GMm}}{{{{\left( R \right)}^2}}} $
Now equating it with the weight of the object it becomes,
$ \Rightarrow \dfrac{{GMm}}{{{{\left( R \right)}^2}}} = mg $
Solving further we get,
$ \Rightarrow \dfrac{{GM}}{{{R^2}}} = g $
Which is the value of the acceleration due to gravity and from the above equation it is clear that $ g $ is independent from the mass of the falling object.
Note
Please note that the distance between the object’s surface is considered negligible, since the radius of the Earth is very large. If the same concept is applied to a large considerable distance, then the value of g will be a different quantity. Also, we can derive from the above activity that whether a soft feather is being thrown down to earth or a heavy metal ball is thrown, the acceleration due to gravity will be the same in both cases.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success
Master Class 11 Computer Science: Engaging Questions & Answers for Success
Master Class 11 Maths: Engaging Questions & Answers for Success
Master Class 11 Social Science: Engaging Questions & Answers for Success
Master Class 11 Economics: Engaging Questions & Answers for Success
Master Class 11 Business Studies: Engaging Questions & Answers for Success
Trending doubts
10 examples of friction in our daily life
What problem did Carter face when he reached the mummy class 11 english CBSE
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
State and prove Bernoullis theorem class 11 physics CBSE
The sequence of spore production in Puccinia wheat class 11 biology CBSE