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Show that the force of gravity between a newborn baby (mass \(=3.0 \mathrm{~kg}\) ) and planet Mars (mass \(=6.4 \times 10^{23} \mathrm{~kg}\) ) is \(4.0 \times 10^{-8} \mathrm{~N}\) when Mars is at its closest to Earth (distance \(=5.6 \times 10^{10} \mathrm{~m}\) ).

Short Answer

Expert verified
The gravitational force between the newborn baby and Mars is indeed \(4.0 \times 10^{-8} \text{N}\).

Step by step solution

01

State the Universal Law of Gravitation

The gravitational force (\(F_g\)) between two objects can be calculated using Newton’s Universal Law of Gravitation: \(F_g = G\frac{m_1 \times m_2}{r^2}\), where \(G\) is the gravitational constant (\(6.674 \times 10^{-11} \text{N}\text{(m/kg)}^2\)), \(m_1\) and \(m_2\) are the masses of the two objects, and \(r\) is the distance between the centers of the two masses.
02

Insert Known Values

Insert the given values into the equation: the mass of the baby (\(m_1 = 3.0 \text{kg}\)), the mass of Mars (\(m_2 = 6.4 \times 10^{23} \text{kg}\)), the distance at closest approach (\(r = 5.6 \times 10^{10} \text{m}\)), and the gravitational constant (\(G = 6.674 \times 10^{-11} \text{N}\text{(m/kg)}^2\)). The formula with the values is then: \(F_g = (6.674 \times 10^{-11} \text{N}\text{(m/kg)}^2)\frac{3.0 \text{kg} \times 6.4 \times 10^{23} \text{kg}}{(5.6 \times 10^{10} \text{m})^2}\).
03

Compute the Gravitational Force

Carry out the calculation: \(F_g = (6.674 \times 10^{-11} \text{N}\text{(m/kg)}^2)\frac{3.0 \text{kg} \times 6.4 \times 10^{23} \text{kg}}{(5.6 \times 10^{10} \text{m})^2} = 4.0 \times 10^{-8} \text{N}\).

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Key Concepts

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

Gravitational Force Calculation
The gravitational force calculation is pivotal for understanding celestial mechanics and the interaction between masses. According to Newton's Universal Law of Gravitation, every pair of objects in the universe attracts each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

The fundamental equation for this calculation is:
\[\begin{equation}F_g = G \frac{m_1 \times m_2}{r^2},\end{equation}\]where:
  • \(F_g\) is the gravitational force between the two masses,
  • \(G\) is the gravitational constant,
  • \(m_1\) and \(m_2\) are the masses of the objects,
  • \(r\) is the distance between the centers of the two masses.
To solve a real-world problem, such as determining the gravitational force between a newborn baby and Mars, you simply substitute the known values into this formula. Precision in measurement is vital, as the gravitational force is often very small and can be significantly affected by even minuscule errors in determining mass or distance. An understanding of the relationships expressed by this equation also enhances comprehension of how gravitational interactions mold the cosmos.
Gravitational Constant
The gravitational constant, denoted by \(G\), is a key part of the gravitational force calculation. Its value is \(6.674 \times 10^{-11} \text{N(m/kg)}^2\), and it is a proportionality factor that makes the gravitational force equation work in all units of measurement.

The gravitational constant was first measured by Henry Cavendish in the late 18th century, a task requiring extreme precision. This constant is crucial because it provides the needed calibration to predict gravitational forces for any two masses at any distance. However, despite its importance, \(G\) is one of the least accurately known constants in physics, with a relatively large uncertainty in its measured value compared to other fundamental constants. Understanding \(G\) can give students a deeper appreciation of how all mass in the universe is interconnected through gravity.
Newton's Laws of Motion
Newton's laws of motion are three physical laws that together laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces.

The first law, also known as the law of inertia, states that an object at rest stays at rest, and an object in motion remains in motion with the same speed and in the same direction unless acted upon by an unbalanced force. The second law defines the relationship between an object's mass, its acceleration, and the applied force, formulated as \(F = ma\). Lastly, the third law states that for every action, there is an equal and opposite reaction.

These laws are deeply connected to the concept of gravity and gravitational forces. For instance, the second law can be utilized to understand how gravitational force causes acceleration in objects. A clear grasp of Newton's laws is essential not only for calculating forces but also for predicting the result of these forces on the motion of objects, which is fundamental in solving a wide array of physics problems.

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Most popular questions from this chapter

Many people mistakenly believe that the astronauts who orbit Earth are "above gravity." Earth's mass is \(6 \times 10^{24} \mathrm{~kg}\), and its radius is \(6.38 \times 10^{6} \mathrm{~m}(6380 \mathrm{~km})\). Use the inverse-square law to show that in "space shuttle territory," 200 kilometers above Earth's surface, the force of gravity on a shuttle is about \(94 \%\) that at Earth's surface.

Satellites are normally sent into orbit by firing them in an easterly direction, the direction in which Earth spins. What is the advantage of this?

Jupiter is more than 300 times as massive as Earth, so it might seem that a body on the surface of Jupiter would weigh 300 times as much as it would weigh on Earth. But it so happens that a body would weigh scarcely three times as much on the surface of Jupiter as it would on the surface of Earth. Discuss why this is so. (Hint: Let the terms in the equation for gravitational force guide your thinking.

A stone is thrown upward at an angle. Neglecting air resistance, what happens to the horizontal component of its velocity along its trajectory?

A penny on its side moving at speed \(v\) slides off the horizontal surface of a table a vertical distance \(y\) from the floor. (a) Show that the penny lands a distance \(v \sqrt{\frac{2 y}{g}}\) from the base of the coffee table. (b) Show that if the speed is \(3.5 \mathrm{~m} / \mathrm{s}\) and the coffee table is \(0.4 \mathrm{~m}\) tall, the distance the coin lands from the base of the table is \(1.0 \mathrm{~m}\). (Use \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\).)

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