Answer
Verified
389.1k+ views
Hint: Using Newton’s law of gravitation find the equation of gravitational constant and acceleration due to gravity. And according to the formula use the relation between acceleration due ro gravity and radius.
Formula used:
$ g=\quad G\dfrac { M }{ { R }_{ E }^{ 2 } }$
Complete step-by-step answer:
Equation of gravitational constant and acceleration due to gravity can be calculated as shown below:
Suppose, a body of mass m is placed on the surface of earth and assumes the shape of the earth to be round. If mass of the earth is M and radius of earth is $ { R }_{ E }$, then by Newton’s law of gravitation we get,
$ F=\quad G\dfrac { Mm }{ { R }_{ E }^{ 2 } }$ …(1)
where, G: Gravitational Constant
Now, from Newton’s Second law of motion,
$ F=\quad mg$ …(2)
where, g: Acceleration due to gravity
From eq.(1) and eq.(2) we get,
$ mg=\quad G\dfrac { Mm }{ { R }_{ E }^{ 2 } }$
$ \therefore \quad g=\quad G\dfrac { M }{ { R }_{ E }^{ 2 } }$
So from the above equation, we can say the value of acceleration due to gravity is least at the maximum radius. Radius is maximum at the equator which means acceleration due to gravity is least at equator.
So, the correct answer is “Option C”.
Note:
As the radius increases acceleration due to gravity decreases. And similarly as the height from the surface of earth increases, acceleration due to gravity decreases. This is given by equation,
$g=\quad G\dfrac { M }{ { ({ R }_{ E }+h) }^{ 2 } } $
where, h: height from the surface of earth
Formula used:
$ g=\quad G\dfrac { M }{ { R }_{ E }^{ 2 } }$
Complete step-by-step answer:
Equation of gravitational constant and acceleration due to gravity can be calculated as shown below:
Suppose, a body of mass m is placed on the surface of earth and assumes the shape of the earth to be round. If mass of the earth is M and radius of earth is $ { R }_{ E }$, then by Newton’s law of gravitation we get,
$ F=\quad G\dfrac { Mm }{ { R }_{ E }^{ 2 } }$ …(1)
where, G: Gravitational Constant
Now, from Newton’s Second law of motion,
$ F=\quad mg$ …(2)
where, g: Acceleration due to gravity
From eq.(1) and eq.(2) we get,
$ mg=\quad G\dfrac { Mm }{ { R }_{ E }^{ 2 } }$
$ \therefore \quad g=\quad G\dfrac { M }{ { R }_{ E }^{ 2 } }$
So from the above equation, we can say the value of acceleration due to gravity is least at the maximum radius. Radius is maximum at the equator which means acceleration due to gravity is least at equator.
So, the correct answer is “Option C”.
Note:
As the radius increases acceleration due to gravity decreases. And similarly as the height from the surface of earth increases, acceleration due to gravity decreases. This is given by equation,
$g=\quad G\dfrac { M }{ { ({ R }_{ E }+h) }^{ 2 } } $
where, h: height from the surface of earth
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE