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Chapter 5: Problem 20

If both the mass and radius of the earth decrease by \(1 \%\), the value of (1) acceleration due to gravity would decrease by nearly \(1 \%\) (2) acceleration due to gravity would increase by \(1 \%\) (3) escape velocity from the earth's surface would decrease by \(1 \%\) (4) the gravitational potential energy of a body on earth's surface will remain unchanged

Short Answer

Expert verified
Acceleration due to gravity increases by about 1%, escape velocity decreases by 1%, and gravitational potential energy remains unchanged.

Step by step solution

01

Understand the Question

We are analyzing the effects on acceleration due to gravity, escape velocity, and gravitational potential energy when both the mass and the radius of Earth decrease by 1%. We must determine which of the given options correctly describes the changes.
02

Formulas for Acceleration Due to Gravity and Escape Velocity

Use the formulas: acceleration due to gravity, \( g = \frac{GM}{R^2} \), and escape velocity, \( v_e = \sqrt{\frac{2GM}{R}} \), where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is its radius.
03

Calculate the New Acceleration Due to Gravity

If both mass \( M \) and radius \( R \) decrease by 1%, the new mass \( M' = 0.99M \) and the new radius \( R' = 0.99R \). The new gravitational acceleration \( g' \) is given by \( g' = \frac{G \cdot 0.99M}{(0.99R)^2} = \frac{0.99G \cdot M}{0.9801R^2} = \frac{G \cdot M}{R^2} \cdot \frac{0.99}{0.9801} \approx 1.01g \). Therefore, the acceleration due to gravity increases by about 1%.
04

Check Effect on Escape Velocity

Using the same decrease, the new escape velocity \( v_e' = \sqrt{\frac{2G \cdot 0.99M}{0.99R}} = \sqrt{\frac{2GM}{R}} \cdot 0.99 = 0.99v_e \). Thus, the escape velocity decreases by 1%.
05

Analyze Gravitational Potential Energy

The gravitational potential energy \( U = -\frac{G M m}{R} \). If both \( M \) and \( R \) decrease by 1%, then \( U' = -\frac{G \cdot 0.99M \cdot m}{0.99R} = U \). Therefore, the gravitational potential energy remains unchanged.

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Key Concepts

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

Escape Velocity
Escape velocity is the speed necessary for an object to break free from the gravitational pull of a celestial body without any additional propulsion. For Earth, this velocity is determined by both its mass and radius. The formula for escape velocity is given by:
  • \( v_e = \sqrt{\frac{2GM}{R}} \)
where:
  • \( G \) is the gravitational constant
  • \( M \) is the mass of the Earth
  • \( R \) is the radius of the Earth
When both Earth's mass and radius decrease by 1%, the escape velocity also decreases. This is calculated as:- New escape velocity, \( v_e' = \sqrt{\frac{2G \cdot 0.99M}{0.99R}} = \sqrt{\frac{2GM}{R}} \cdot 0.99 = 0.99v_e \)Thus, the escape velocity decreases by 1%. This means objects would have a slightly easier time escaping Earth's gravity under these conditions.
Gravitational Potential Energy
Gravitational potential energy represents the energy an object posseses due to its position in a gravitational field. For an object at Earth's surface, it is described by the formula:
  • \( U = -\frac{GMm}{R} \)
where:
  • \( G \) is the gravitational constant
  • \( M \) is the mass of the Earth
  • \( m \) is the mass of the object
  • \( R \) is the Earth's radius
If both Earth's mass and radius are reduced by 1%, the expression for gravitational potential energy becomes:- New gravitational potential energy \( U' = -\frac{G \cdot 0.99M \cdot m}{0.99R} \)This simplifies back to \( U \), meaning the gravitational potential energy remains unchanged. Simply put, despite changes in mass and radius, the potential for an object to do work by virtue of its position does not shift.
Changes in Earth's Mass and Radius
Alterations in Earth's mass and radius significantly impact gravitational forces and related quantities. Let's explore these changes:1. **Impact on Gravity**: Gravity at the surface of the Earth is governed by: - \( g = \frac{GM}{R^2} \) When mass and radius decrease by 1%, the new gravitational force is slightly stronger. Calculations show: - New gravity \( g' = \frac{0.99GM}{(0.99R)^2} \approx 1.01g \) This results in an increase of about 1% in gravitational acceleration, meaning things weigh slightly more. 2. **Impact on Escape Velocity**: With a smaller mass and radius, escape velocity falls, as discussed earlier. This change suggests lesser effort is needed to overcome Earth's gravity.These changes illustrate how sensitive gravitational calculations can be and how even small changes in fundamental properties, like Earth's size and mass, influence our physical environment.

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Most popular questions from this chapter

Which of the following are correct? (1) If \(R\) is the radius of a planet, \(g\) is the acceleration due to gravity, the mean density of the planet is \(3 g / 4 \pi G R\). (2) Acceleration due to gravity is a universal constant. (3) The escape velocity from a planet which has double the mass of earth and half its radius is \(22.4 \mathrm{~km} \mathrm{~s}^{-1}\) (4) The ratio of gravitational mass and inertial mass of a body at the surface of earth is 1 .

Four similar particles of mass \(m\) are orbiting in a cirt? of radius \(r\) in the same angular direction because of thir mutual gravitational attractive force. Velocity of a particle is given by (1) \(\left[\frac{G m}{r}\left(\frac{1+2 \sqrt{2}}{4}\right)\right]^{\frac{1}{2}}\) (2) \(\sqrt{\frac{G m}{r}}\) (3) \(\sqrt{\frac{G m}{r}(1+2 \sqrt{2})}\) (4) \(\left[\frac{1}{2} \frac{G m}{r}\left(\frac{1+2 \sqrt{2}}{2}\right)\right]^{1}\)

The earth is assumed to be a sphere of radius \(R\). A platform is arranged at a height \(3 R\) from the surface of the earth. The escape velocity of a body from this platform is \(f v_{e}\), where \(v_{e}\) is its escape velocity from the surface of the earth. Find the value of \(f\).

Consider a planet moving in an elliptical orbit around the Sun. The work done on the planet by the gravitational force of the Sun (1) is zero in any small part of the orbit (2) is zero in some parts of the orbit (3) is zero in complete revolution (4) is zero in no part of the motion

\- A body is thrown from the surface of the earth with velocity \(\left(g R_{e}\right) / 2\), where \(R_{e}\) is the radius of the earth at some angle from the vertical. If the maximum height reached by the body is \(R / 4\), then the angle of projection with the vertical is (1) \(\sin ^{-1}\left(\frac{\sqrt{5}}{4}\right)\) (2) \(\cos ^{-1}\left(\frac{\sqrt{5}}{4}\right)\) (3) \(\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\) (4) none of these
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