What is value of gravitational acceleration at the center of the earth ?

9.8 m s^{2}

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6.67 \times 10^{- 11} m s^{- 2}

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Solution

The correct option is A $9.8 m s^{2}$
Its value is 9.8 m/s2 on Earth. That is to say, the acceleration of gravity on the surface of the earth at sea level is 9.8 m/s2. the acceleration of gravity, it was mentioned that the value of g is dependent upon location. There are slight variations in the value of g about earth's surface.

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Similar questions

Q. What will be change in value of gravitational acceleration with increase in r from the center of the earth?

Q. The value of gravitational acceleration g at the center of the earth is:-

Q. Solve:
Estimate the mass of the earth, given, radius of the earth

= 6.4 \times 10^{6} m

, acceleration due to gravity

= 9.8 m / s^{2}

and gravitational constant

= 6.67 \times 10^{- 11} S . I .

units.

Q. Radius of earth is equal to

6 \times 10^{6} m

. Acceleration due to gravity is equal to

9.8 m / s^{2}

. Gravitational constant

G

is equal to

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

. Then mass of the earth is:-

Two spherical balls of mass 10 kg each are placed with their centers 10 cm apart. Find the gravitational force of attraction between them. Take $G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}$

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