The average value of acceleration due to gravity on the surface of the earth is equal to :

9.8 ms

^{- 2}

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9.8 ms

^{- 1}

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19.6 ms

^{- 2}

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9.8 s

^{- 2}

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Solution

The correct option is A 9.8 ms $^{- 2}$
Acceleration due to the gravity at earth $= g = \frac{G M}{R^{2}}$
Where G=Gravitational constant $= 6.673 \times 10^{- 11}$
M=Mass of earth $= 5.972 \times 10^{24} k g$ and
R=Radius of earth $= 6.371 \times 10^{6} m$

Putting these value in the above formula, we will get $g = 9.8 m / s^{2}$

Hence correct answer is option $A$

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Similar questions

Q. If mass and radius of the Earth are

6.0 \times 10^{24} k g

and

6.4 \times 10^{6} m

respectively. Calculate the force exerted by the Earth on a body of mass 1 kg, acceleration of the body of mass 1 kg and acceleration of the Earth in order.
Use

G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}

What is the value of acceleration due to gravity of the Earth?
(Given mass of the Earth is $6 \times 10^{24} k g$ and radius of the Earth is $6.4 \times$ $10^{6} m$ ).

Q. A person weighs

110.84 N

on the moon whose acceleration due to gravity is one-sixth of that earth if the value of g on earth is

9.8 m s^{- 2}

calculate, acceleration due to gravity on the earth.

Q. If the mass of a body is

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its weight on the surface of the earth would be ___. [Take acceleration due to gravity,

g = 9.8 m s^{- 2}]

The acceleration due to gravity on the surface of moon is 1.7 ms^–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 ms^–2)

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The average value of acceleration due to gravity on the surface of the earth is equal to :

The correct option is A 9.8 ms−2Acceleration due to the gravity at earth=g=GMR2Where G=Gravitational constant =6.673×10−11M=Mass of earth =5.972×1024kg andR=Radius of earth =6.371×106mPutting these value in the above formula, we will get g=9.8m/s2Hence correct answer is option A