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(I) Calculate the acceleration due to gravity on the Moon. The Moon's radius is \(1.74 \times 10^{6} \mathrm{~m}\) and its mass is \(7.35 \times 10^{22} \mathrm{~kg}\).

Short Answer

Expert verified
The gravitational acceleration on the Moon is approximately 1.62 m/s².

Step by step solution

01

Understand the Formula for Gravitational Acceleration

To find the acceleration due to gravity on the Moon, you can use the formula: \[ g = \frac{{G imes M}}{{R^2}} \]where \(g\) is the gravitational acceleration, \(G\) is the gravitational constant \((6.674 \times 10^{-11} \, \text{N(m/kg)}^2)\), \(M\) is the mass of the Moon, and \(R\) is the radius of the Moon.
02

Input Known Values into the Formula

Substitute the known values into the formula:\[ g = \frac{{6.674 \times 10^{-11} \, \text{N(m/kg)}^2 \times 7.35 \times 10^{22} \, \text{kg}}}{{(1.74 \times 10^6 \, \text{m})^2}} \]
03

Calculate the Denominator

Calculate the square of the Moon's radius:\[ R^2 = (1.74 \times 10^6)^2 = 3.0276 \times 10^{12} \, \text{m}^2 \]
04

Calculate the Numerator

Multiply the gravitational constant \(G\) and the mass \(M\) of the Moon:\[ 6.674 \times 10^{-11} \, \text{N(m/kg)}^2 \times 7.35 \times 10^{22} \, \text{kg} = 4.90239 \times 10^{12} \, \text{Nm}^2/\text{kg} \]
05

Divide Numerator by Denominator

Use the values from Step 3 and Step 4 to find \(g\):\[ g = \frac{{4.90239 \times 10^{12} \, \text{Nm}^2/\text{kg}}}{{3.0276 \times 10^{12} \, \text{m}^2}} = 1.62 \, \text{m/s}^2 \]
06

Conclusion

The acceleration due to gravity on the Moon is calculated to be approximately \(1.62 \, \text{m/s}^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Constant
The gravitational constant, denoted by \(G\), plays a crucial role in understanding gravitational forces in our universe. It is effectively the "gluing" factor that allows us to calculate the force of gravity between any two masses.
\(G\) has a specific value of \(6.674 \times 10^{-11} \, \text{N(m/kg)}^2\). This small constant reflects the weak nature of gravitational forces compared to other fundamental forces in nature, such as electromagnetic forces.
To comprehend how \(G\) is applied, consider it as the connection between mass and gravitational pull. It ensures that when calculating gravity, we account for the size of the masses involved and the distance between them.
This constant remains unchanged across various environments, including the Earth and the Moon, making it invaluable for calculating gravitational acceleration in different celestial locations.
Mass of the Moon
The mass of the Moon is a key factor in determining the gravitational acceleration on its surface. With a mass of \(7.35 \times 10^{22} \, \text{kg}\), the Moon is much less massive than Earth.
This difference in mass accounts for the Moon's weaker gravitational pull, which is about one-sixth that of Earth's gravity.
Understanding the Moon's mass is essential for calculations involving space travel, lunar exploration, and astronomy. When we speak of mass in physics, we're referring to the amount of matter in an object—not its weight.
In the absence of other forces, the mass of an astronomical body directly influences the strength of its gravitational field. This is why astronauts on the Moon can jump higher and carry heavy equipment with less effort.
Radius of the Moon
The Moon's radius is another critical variable when calculating gravitational acceleration on its surface. Measured to be \(1.74 \times 10^{6} \, \text{m}\), the radius directly affects how strong or weak the gravity will be.
The gravitational pull experienced at the surface of the Moon is inversely proportional to the square of its radius.
This relationship is fundamental in physics, as it shows how gravitational force decreases with distance. The law of inverse squares states that as you move away from a massive body like the Moon, the pull of its gravity reduces rapidly.
When engaging in problems relating to gravity and celestial objects, always consider both the mass and radius of the object.
  • The smaller the radius, the stronger the surface gravity for a fixed mass.
  • The larger the radius, the weaker the surface gravity, assuming mass remains constant.
This concept helps scientists and engineers design efficient spacecraft trajectories and plan lunar missions.

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Most popular questions from this chapter

(II) Two satellites orbit Earth at altitudes of \(5000 \mathrm{~km}\) and \(15,000 \mathrm{~km} .\) Which satellite is faster, and by what factor?

(II) What is the apparent weight of a \(75-\mathrm{kg}\) astronaut \(2500 \mathrm{~km}\) from the center of the Earth's Moon in a space vehicle \((a)\) moving at constant velocity and \((b)\) accelerating toward the Moon at \(2.3 \mathrm{~m} / \mathrm{s}^{2}\) ? State "direction" in each case.

(II) Calculate the period of a satellite orbiting the Moon, \(120 \mathrm{~km}\) above the Moon's surface. Ignore effects of the Earth. The radius of the Moon is \(1740 \mathrm{~km}\).

The Sun rotates about the center of the Milky Way Galaxy (Fig. 29\()\) at a distance of about \(30,000\) light-years from the center \(\left(11 y=9.5 \times 10^{15} \mathrm{m}\right) .\) If it takes about 200 million years to make one rotation, estimate the mass of our Galaxy. Assume that the mass distribution of our Galaxy is concentrated mostly in a central uniform sphere. If all the stars had about the mass of our Sun \(\left(2 \times 10^{30} \mathrm{kg}\right),\) how many stars would there be in our Galaxy?

Estimate the value of the gravitational constant \(G\) in Newton's law of universal gravitation using the following data: the acceleration due to gravity at the Earth's surface is about \(10 \mathrm{~m} / \mathrm{s}^{2} ;\) the Earth has a circumference of about \(40 \times 10^{6} \mathrm{~m}\); rocks found on the Earth's surface typically have densities of about \(3000 \mathrm{~kg} / \mathrm{m}^{3}\) and assume this density is constant throughout (even though you suspect it is not true).

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