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Chapter 6: Problem 28

If an object's weight is \(W\) on the earth, what would be its weight (in terms of \(W )\) if the earth had (a) twice its present mass, but was the same size, (b) half its present radius, but the same mass, (c) half its present radius and half its present mass, (d) twice its present radius and twice its present mass?

Short Answer

Expert verified
(a) 2W, (b) 4W, (c) 2W, (d) 0.5W.

Step by step solution

01

Understanding the Weight Formula

Weight on Earth is given by the formula \( W = mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity. However, \( g \) can also be expressed as \( g = \frac{GM}{R^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.
02

Part (a): Effect of Doubling Earth's Mass

When the Earth's mass is doubled, the new gravitational acceleration \( g' \) becomes \( g' = \frac{2GM}{R^2} = 2g \). This results in the new weight \( W' = mg' = m(2g) = 2W \). Thus, the object's weight would be twice its original weight, \( 2W \).
03

Part (b): Effect of Halving Earth's Radius

If Earth's radius is halved, the new gravitational acceleration \( g'' \) is \( g'' = \frac{GM}{(R/2)^2} = \frac{GM}{R^2/4} = 4g \). So, \( W'' = mg'' = m(4g) = 4W \). The object's weight would be four times its original weight, \( 4W \).
04

Part (c): Halving Both Radius and Mass

When both radius and mass are halved, the gravitational acceleration \( g''' \) becomes \( g''' = \frac{G(M/2)}{(R/2)^2} = \frac{G(M/2)}{R^2/4} = 2g \). Thus, \( W''' = m(2g) = 2W \). The object's weight would be twice its original weight, \( 2W \).
05

Part (d): Doubling Both Radius and Mass

If both the radius and mass are doubled, the new gravitational acceleration \( g'''' \) is \( g'''' = \frac{G(2M)}{(2R)^2} = \frac{G(2M)}{4R^2} = \frac{1}{2}g \). Hence, \( W'''' = m\left(\frac{1}{2}g\right) = \frac{1}{2}W \). Therefore, the object's weight would be half its original weight, \( \frac{1}{2}W \).

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Key Concepts

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

Acceleration Due to Gravity
The acceleration due to gravity, commonly represented by the symbol \( g \), is a measure of the force of gravity acting on an object at the surface of a large body such as Earth. This is the constant rate at which an object accelerates when free-falling towards Earth's surface without any resistance. It generally rounds off to \( 9.81 \, \text{m/s}^2 \) on Earth. The formula used to calculate \( g \) is:
  • \( g = \frac{GM}{R^2} \)
Where:
  • \( G \) is the gravitational constant \( \approx 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)
  • \( M \) is the mass of the Earth
  • \( R \) is the radius of the Earth
Understanding this formula is crucial because it shows how \( g \) changes when either \( M \) (the mass of the Earth) or \( R \) (the radius of the Earth) is altered, which directly affects an object's weight.
Mass of the Earth
The mass of the Earth is a fundamental component in determining the gravitational force exerted by the planet. This mass is roughly estimated to be \( 5.972 \times 10^{24} \, \text{kg} \). It plays a key role in our formula for gravitational force and acceleration due to gravity. When the Earth's mass changes, it impacts the gravitational pull:
  • Doubling Earth's mass results in a doubled gravitational force.
  • Halving Earth's mass leads to halved gravitational strength.
These changes are reflected in shifts in the weight of objects, demonstrating the relationship between mass and gravity.
Radius of the Earth
The radius of the Earth is the distance from Earth's center to its surface and is crucial in calculating gravitational force. The average radius is approximately \( 6,371 \, \text{km} \). Adjustments in this radius directly influence gravitational force and, therefore, object weight:
  • Halving the radius increases Earth's gravitational pull fourfold.
  • Doubling the radius decreases Earth's gravitational effect by a factor of four.
This is because gravitational acceleration is inversely proportional to the square of the radius \( (R^2) \), meaning any change to \( R \) has a significant impact. These principles help to understand how the weight of an object would change with varying sizes of planets.
Weight Formula
The weight of an object on Earth is calculated using the weight formula \( W = mg \). In this formula:
  • \( m \) is the mass of the object, a constant in any local environment.
  • \( g \) is the acceleration due to gravity, which can change depending on the mass and radius of Earth or another celestial body.
This formula highlights that weight is not the same as mass. Instead, it is the force exerted on the mass due to gravity. Consequently, changes to \( g \) from varying conditions (like the ones outlined in the task - changes to Earth's mass and radius) lead to changes in weight. Understanding this relationship is key to solving problems dealing with gravitational force and object weight on different planets or under different conditions.

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Most popular questions from this chapter

A bowling ball weighing 71.2 \(\mathrm{N}\) is attached to the ceiling by a 3.80 \(\mathrm{m}\) rope. The ball is pulled to one side and released; it then swings back and forth like a pendulum. As the rope swings through its lowest point, the speed of the bowling ball is measured at 4.20 \(\mathrm{m} / \mathrm{s}\) . At that instant, find (a) the magnitude and direction of the acceleration of the bowling ball and (b) the tension in the rope. Be sure to start with a free-body diagram.

Force on a skater's wrist. A 52 \(\mathrm{kg}\) ice skater spins about a vertical axis through her body with her arms horizontally out-stretched, making 2.0 turns each second. The distance from one hand to the other is 1.50 \(\mathrm{m} .\) Biometric measurements indicate that each hand typically makes up about 1.25\(\%\) of body weight. (a) Draw a free-body diagram of one of her hands. (b) What horizontal force must her wrist exert on her hand? (c) Express the force in part (b) as a multiple of the weight of her hand.

The star Rho \(^{1}\) Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a circular orbit around Rho I Cancri with an orbital radius equal to 0.11 times the radius of the earth's orbit around the sun. What are (a) the orbital speed and (b) the orbital period of the planet of Rho "Cancri?

Stunt pilots and fighter pilots who fly at high speeds in a downward-curving arc may experience a "red out," in which blood is forced upward into the flier's head, potentially swelling or breaking capillaries in the eyes and leading to a reddening of vision and even loss of consciousness. This effect can occur at centripetal accelerations of about 2.5\(g^{\prime}\) s. For a stunt plane flying at a speed of 320 \(\mathrm{km} / \mathrm{h}\) , what is the minimum radius of downward curve a pilot can achieve without experiencing a red out at the top of the arc? (Hint: Remember that gravity provides part of the centripetal acceleration at the top of the arc; it's the acceleration required in excess of gravity that causes this problem.)

A stone with a mass of 0.80 \(\mathrm{kg}\) is attached to one end of a string 0.90 \(\mathrm{m}\) long. The string will break if its tension exceeds 60.0 \(\mathrm{N} .\) The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed. (a) Make a free-body diagram of the stone. (b) Find the maximum speed the stone can attain without breaking the string.
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