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
[tex]\texttt{ }[/tex]
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!
[tex]\texttt{ }[/tex]
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]
[tex]\texttt{ }[/tex]
Conclusion:
Acceleration of gravity at the surface of Mercury is about 3.8 m/s²
[tex]\texttt{ }[/tex]
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
[tex]\texttt{ }[/tex]
Answer details
Grade: High School
Subject: Mathematics
Chapter: Gravitational Force
