Question

9. The acceleration due to gravity near earth's surface is
9.8m/s2 and the radius of earth is 6400km. Calculate the
mass of earth using the above information and G's value.​

Answer 1

Answer:

6.63×10^-24

Explanation:

g=GM/R2

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Answer 2

Given :

• Radius of the Earth = 6400 km / 6.4 × 10⁶m.

• Acceleration due to Gravity = 9.8 m/s²

• G = 6.67 × 10¯¹¹ N m²kg¯²

To Find :-

The Mass of the Earth.

Solution :-

Let the mass of earth be m kg.

We know the formula for acceleration due to gravity on an planet !! i.e,

$\underline{\boxed{\bf{g = \dfrac{GM}{r^{2}}}}}$

Where :-

• g = Acceleration due to gravity
• M = Mass of the planet
• r = Distance

Now using the formula and substituting the values in it, we get :-

$:\implies \bf{9.8 = \dfrac{6.67 \times 10^{-11} \times M}{(6.4 \times 10^{6})^{2}}}$

$:\implies \bf{9.8 = \dfrac{6.67 \times \times 10^{-11} \times M}{(6.4 \times 10^{6})(6.4 \times 10^{6})}}$

$:\implies \bf{9.8 = \dfrac{6.67 \times 10^{-11} \times M}{6.4 \times 6.4 \times 10^{12}}}$

$:\implies \bf{9.8 = \dfrac{6.67 \times 10^{-11} \times M}{40.96 \times 10^{12}}}$

Now , multiplying both the sides by (40.96 × 10¹²) , we get :-

$:\implies \bf{9.8 \times 40.96 \times 10^{12} = \dfrac{6.67 \times 10^{-11} \times M}{40.96 \times 10^{12}} \times 40.96 \times 10^{12}}$

$:\implies \bf{401.4 \times 10^{12} = 6.67 \times 10^{-11} \times M}$

$:\implies \bf{\dfrac{401.4 \times 10^{12}}{6.67} = 10^{-11} \times M}$

$:\implies \bf{60(approx.) \times 10^{12} = 10^{-11} \times M}$

$:\implies \bf{\dfrac{60 \times 10^{12}}{10^{-11}} = M}$

$:\implies \bf{60 \times 10^{12 + 11} = M}$

$:\implies \bf{60 \times 10^{23} = M}$

$:\implies \bf{6 \times 10^{24} = M}$

$\therefore \bf{M = 6 \times 10^{24}}$

Hence, the mass of earth is 6 × 10²⁴ kg.