Question

Calculate the value of acceleration due to gravity on the surface of the mars. (Mass of mars =6.4×10^23 radius =3400 km).

Answer 1

$<b><u>Heya mate!! Here's your solution</u>$
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$\huge{Given}$

Mass of Mars is 6.4 × 10^23 and it's radius is 3400 km.

$\huge{To\: Find}$

Acceleration due to gravity on surface of Mars.

$\huge{Solution}$

Mass of Mars (M) = 6.4 × 10^23

Radius of Mars (R)
= 3400 km
= 3400000 m
= 3.4 × 10^6 m

And, the universal gravitational constant (G) = 6.7 × 10^-11
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Let the acceleration due to gravity on Mars be g.

Now, we know that,

F = mg [second law of motion]

Also,

F = GMm/d² [universal gravitational law]

Then,

mg = GMm/d²

=> g = GM/d²

=> g = GM/R²

$= > g = \frac{6.7 \times {10}^{ - 11} \times 6.4 \times {10}^{23} }{(3.4 \times {10}^{6} ) ^{2} }$

$= > g = \frac{6.7 \times 6.4 \times {10}^{ - 11} \times {10}^{23} }{3.4 \times 3.4 \times {10}^{6} \times {10}^{6} }$

$= > g = \frac{6.7 \times 6.4 \times {10}^{ - 11 + 23 - 6 - 6} }{3.4 \times 3.4}$

$= > g = \frac{6.7 \times 6.4 \times {10}^{0} }{3.4 \times 3.4}$

$= > g = \frac{42.88×1}{11.56}$

$= > g = 3.709$

Hence, acceleration due to gravity (g) on Mars is 3.7 m/s².
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$\mathcal{\bold{\blue{Finally,}}}$

Your answer is -

$\boxed{\bold{\mathcal{\red{3.7\: m/s^2}}}}$
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$\huge{\bold{\mathfrak{\green{Thank\: you}}}}$

Answer 2

Hey here's your answer

Plz mark as brainlist