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Chapter 6: Problem 697

If the mass of earth is 80 times of that of a planet and diameter is double that of planet and ' \(\mathrm{g}\) ' on the earth is \(9.8 \mathrm{~ms}^{-2}\), then the value of \(\mathrm{g}^{\prime}\) on that planet is \(=\ldots \ldots \ldots \mathrm{ms}^{-2}\) (A) \(4.9\) (B) \(0.98\) (C) \(0.49\) (D) 49

Short Answer

Expert verified
The value of $g'$ on the planet is \(4.9 \mathrm{~ms}^{-2}\). The correct choice is (A).

Step by step solution

01

Define the given quantities

We are given: 1. Mass of Earth is 80 times the mass of the planet: \(M_{Earth} = 80 M_{Planet}\) 2. Diameter of Earth is 2 times the diameter of the planet: \(D_{Earth} = 2D_{Planet}\) 3. g on Earth = \(9.8 \mathrm{~ms}^{-2}\)
02

Find the relationship between g and mass

The formula for gravitational force is: \(F = \cfrac{GMm}{r^{2}}\) Where: F is the gravitational force between two masses, G is the gravitational constant (\(6.674 \times 10^{-11} \mathrm{Nm^{2}kg^{-2}}\)), M and m are the two masses, r is the distance between their centers. The force acting on an object due to gravity is the gravitational force, so using F = ma, \(ma = \cfrac{GMm}{r^{2}}\) Here, a is the acceleration due to gravity (g), so: \(g = \cfrac{GM}{r^{2}}\) The value of g depends on the mass of the planet and the distance from its center (which is half of the diameter, i.e., its radius).
03

Solve for g' on the planet

Let g' be the acceleration due to gravity on the planet. Using the given data, we can rewrite the relationship found earlier: \(g' = \cfrac{G(80 M_{Planet})}{(2r_{planet})^{2}}\) Simplify the equation: \(g' = \cfrac{80 GM_{Planet}}{4r_{planet}^{2}}\) \(g' = \cfrac{20 GM_{Planet}}{r_{planet}^{2}}\) Now, we have the relationship \(g_{Earth} = \cfrac{GM_{Earth}}{r_{Earth}^{2}}\) and since \(g_{Earth} = 9.8 \mathrm{~ms}^{-2}\), we can plug the values. \(9.8 = \cfrac{G(80 M_{Planet})}{r_{Earth}^{2}}\) Dividing the two equations: \(\cfrac{g'}{9.8} = \cfrac{\cfrac{20 GM_{Planet}}{r_{planet}^{2}}}{\cfrac{80 GM_{Planet}}{r_{Earth}^{2}}}\) Cancel terms: \(\cfrac{g'}{9.8} = \cfrac{20}{80}\cfrac{r_{Earth}^{2}}{r_{planet}^{2}}\) Given relationship: \(D_{Earth} = 2D_{Planet}\), so \(r_{Earth} = 2 r_{Planet}\) \(\cfrac{g'}{9.8} = \cfrac{20}{80}\left(\cfrac{2r_{planet}}{r_{planet}}\right)^{2}\) \(\cfrac{g'}{9.8} = \cfrac{20}{80} (2^{2})\) \(\cfrac{g'}{9.8} = \cfrac{1}{2}\) Finally, solve for g': \(g' = 9.8 \times \cfrac{1}{2}\) \(g' = 4.9 \mathrm{~ms^{-2}}\)
04

Compare with the given choices

The calculated value of g' on the planet is \(4.9 \mathrm{~ms}^{-2}\), which matches option (A). Therefore, the correct choice is (A) \(4.9 \mathrm{~ms}^{-2}\).

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Key Concepts

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

Mass and Gravitational Acceleration
In the context of gravitational acceleration, mass is an essential factor to consider. The gravitational force that an object exerts depends on its mass. Larger masses exert more gravitational force. For example, the Earth has a substantial mass, which results in a gravitational acceleration of approximately \(9.8 \mathrm{~ms}^{-2}\). The formula for gravitational acceleration \(g\) is given by:\[ g = \frac{GM}{r^2} \]Here:
  • \(G\) is a gravitational constant.
  • \(M\) is the mass of the object, such as a planet.
  • \(r\) is the distance from the center of the mass to the point of measurement.
When comparing two bodies like Earth and a planet, the gravitational acceleration experienced on a planet with a mass \(M_{Planet}\) much smaller than Earth's would obviously have a lower gravitational pull.
Radius and Its Effect
The radius of a planet significantly affects gravitational acceleration as it represents the distance over which the planet's gravity acts. The larger the radius, the weaker the gravitational acceleration at the surface because the force spreads over a larger area. The fundamental relation, again, is \(g = \cfrac{GM}{r^{2}}\), indicating that g is inversely proportional to the square of the radius (\(r^2\)). This means even a slight increase in the radius results in a significant decrease in gravitational acceleration. In our exercise, the diameter (and thus the radius) of Earth is given as twice that of the planet, which alters the value of \(r\) in the equation inversely, thereby affecting the gravitational acceleration greatly. This shows why understanding a planet's radius is key when we're investigating how strong its gravity will feel.
Gravitational Constant and Its Role
The gravitational constant \(G\) is a universal constant and is instrumental in the calculations involving gravity. Its value is \(6.674 \times 10^{-11} \mathrm{Nm^{2}kg^{-2}}\). It provides the framework that allows us to calculate the gravitational pull between two masses; without \(G\), we wouldn't be able to calculate gravitational forces accurately.\(G\) is vital in our equations:
  • \(F = \cfrac{GMm}{r^{2}}\) for gravitational force,
  • \(g = \cfrac{GM}{r^{2}}\) for gravitational acceleration.
These formulas show that \(G\) acts consistently across calculations, making it essential for understanding gravity not only on Earth but in celestial mechanics across different planets and systems.
Planetary Comparison
Comparing gravity on different planets involves looking at both mass and radius. With the exercise given, the problem highlights how changes in these quantities affect the gravity reading:
  • The Earth is much more massive, having 80 times the mass of the planet compared.
  • Its diameter is twice the planet's, which directly influences the gravitational pull felt on its surface.
  • The acceleration due to gravity \(g'\) on the planet is calculated to be half of Earth's gravity value as a result of mass and radius variations.
This kind of analysis allows scientists to estimate not just gravitational forces but also how surface conditions, like weight that can be felt, would differ across celestial bodies. Such comparisons are foundational in astronomy and space exploration, helping us understand not just our planet but the wider universe.

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Most popular questions from this chapter

A satellite revolves around the earth in an elliptical orbit. Its speed (A) is the same at all points in the orbit (B) is greatest when it is closest to the earth (C) is greatest when it is farthest to the earth (D) goes on increasing or decreasing continuously depending upon the mass of the satellite

The escape velocity of a body from earth's surface is Ve. The escape velocity of the same body from a height equal to 7 R from earth's surface will be (A) \((\mathrm{Ve} / \sqrt{2})\) (B) \((\mathrm{Ve} / 2)\) (C) \((\mathrm{Ve} / 2 \sqrt{2})\) (D) \((\mathrm{Ve} / 4)\)

Two satellites of mass \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\left(\mathrm{~m}_{1}=\mathrm{m}_{2}\right)\) are revolving round the earth in circular orbits of \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\left(\mathrm{r}_{1}>\mathrm{r}_{2}\right)\) respectively. Which of the following statement is true regarding their speeds \(\mathrm{V}_{1}\) and \(\mathrm{V}_{2}\) (A) \(\mathrm{V}_{1}=\mathrm{V}_{2}\) (B) \(\mathrm{V}_{1}<\mathrm{V}_{2}\) (C) \(\mathrm{V}_{1}>\mathrm{V}_{2}\) (D) \(\left(\mathrm{V}_{1} / \mathrm{r}_{1}\right)=\left(\mathrm{V}_{2} / \mathrm{r}_{2}\right)\)

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If the radius of the earth were to shrink by \(1 \%\) its mass remaining the same, the acceleration due to gravity on the earth's surface would (A) decrease by \(2 \%\) (B) remain Unchanged (C) increase by \(2 \%\) (D) increases by \(1 \%\)
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