Respuesta :

Answer:

[tex]3.69 m/s^2[/tex]

Explanation:

The acceleration due to gravity on the surface of a planet is given by:

[tex]g=\frac{GM}{R^2}[/tex]

where:

[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}[/tex] is the gravitational constant

M is the mass of the planet

R is the radius of the planet

For Mercury, we have:

[tex]M=3.29\cdot 10^{23} kg[/tex] is the mass

[tex]R=2,440 km=2.44\cdot 10^6 m[/tex] is the radius

So, the acceleration due to gravity on the surface is

[tex]g=\frac{(6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2})(3.29\cdot 10^{23} kg)}{(2.44\cdot 10^6 m)^2}=3.69 m/s^2[/tex]

Acceleration of gravity at the surface of Mercury is about 3.8 m/s²

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Further explanation

Let's recall the Gravitational Force formula:

[tex]\boxed {F = G\ \frac{m_1 m_2}{R^2}}[/tex]

where:

F = Gravitational Force ( N )

G = Gravitational Constant ( = 6.67 × 10⁻¹¹ Nm²/kg² )

m = mass of object ( kg )

R = distance between object ( m )

Let us now tackle the problem!

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Given:

mass of Mercury = M = 3.3 × 10²³ kg

radius of Mercury = R = 2400 km = 2.4 × 10⁶ m

Asked:

acceleration of gravity at the surface of Mercury = g = ?

Solution:

We will calculate the acceleration of gravity at the surface of Mercury by using following formula:

[tex]F = G\ \frac{M m}{R^2}[/tex]

[tex]\frac{ F }{ m} = G\frac{M}{R^2}[/tex]

[tex]g = G\frac{M}{R^2}[/tex]

[tex]g = 6.67 \times 10^{-11} \frac{3.3 \times 10^{23}}{( 2.4 \times 10^6)^2}[/tex]

[tex]g \approx 3.8 \texttt{ m/s}^2[/tex]

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Conclusion:

Acceleration of gravity at the surface of Mercury is about 3.8 m/s²

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Learn more

  • Unit of G : https://brainly.com/question/1724648
  • Velocity of Runner : https://brainly.com/question/3813437
  • Kinetic Energy : https://brainly.com/question/692781
  • Acceleration : https://brainly.com/question/2283922
  • The Speed of Car : https://brainly.com/question/568302

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Answer details

Grade: High School

Subject: Mathematics

Chapter: Gravitational Force

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