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Gravity is a fundamental natural force that pulls objects toward each other. It is the reason why things fall to the ground when dropped and why the Earth orbits the Sun. Although we experience gravity every day, its true nature is universal-it acts between any two masses anywhere in the universe.
Historically, gravity was first described by Sir Isaac Newton in the 17th century, who formulated the law explaining how every mass attracts every other mass. Understanding gravity helps us explain phenomena ranging from why apples fall to the ground to how planets move in space.
In this chapter, we will explore the concept of gravity from its basic principles, learn the mathematical laws governing it, and see how it affects objects on Earth and beyond.
Imagine you and a friend standing a few meters apart. Although you cannot feel it, there is a tiny gravitational pull between you and your friend because both of you have mass. This attraction is called gravitational force.
Newton's law of gravitation states that every two masses attract each other with a force that is:
This means that the heavier the objects, the stronger the pull, and the farther apart they are, the weaker the pull.
Mathematically, Newton's law of gravitation is written as:
Here, G is a very small constant with the value \(6.674 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2\). This small value explains why gravitational force between everyday objects is tiny and usually unnoticeable.
When you drop an object, it accelerates toward the Earth. This acceleration is caused by Earth's gravitational pull and is called acceleration due to gravity, denoted by g.
On average, near Earth's surface, the value of g is approximately \(9.8 \, \text{m/s}^2\). This means that every second, the speed of a falling object increases by 9.8 meters per second, assuming air resistance is negligible.
Earth itself is a massive object with mass \(M\) and radius \(R\). Using Newton's law, the acceleration due to gravity at the surface is:
The value of g is not constant everywhere. It changes depending on how far you are from Earth's center.
At height \(h\) above the surface:
At depth \(d\) below the surface:
Two terms often confused are mass and weight. Understanding their difference is crucial.
| Property | Mass | Weight |
|---|---|---|
| Definition | Amount of matter in an object | Force due to gravity on the object |
| Symbol | m | W |
| Units | Kilogram (kg) | Newton (N) |
| Value | Constant everywhere | Varies with location (depends on g) |
| Formula | - | \( W = mg \) |
Gravity influences many phenomena we observe:
In physics, we use the SI system for all measurements:
Using consistent units is important to avoid errors in calculations involving gravity.
Gravity plays a vital role in many areas:
Step 1: Identify the given values:
Step 2: Use Newton's law of gravitation formula:
\[ F = G \frac{m_1 m_2}{r^2} = 6.674 \times 10^{-11} \times \frac{5 \times 5}{2^2} \]Step 3: Calculate numerator and denominator:
\[ 5 \times 5 = 25, \quad 2^2 = 4 \]Step 4: Substitute values:
\[ F = 6.674 \times 10^{-11} \times \frac{25}{4} = 6.674 \times 10^{-11} \times 6.25 = 4.17125 \times 10^{-10} \, \text{N} \]Answer: The gravitational force between the two masses is approximately \(4.17 \times 10^{-10} \, \text{N}\), which is extremely small.
Step 1: Identify the given values:
Step 2: Use the formula for gravity at height \(h\):
\[ g_h = g \left(1 - \frac{h}{R}\right)^2 \]Step 3: Calculate \(\frac{h}{R}\):
\[ \frac{1000}{6.4 \times 10^6} = 1.5625 \times 10^{-4} \]Step 4: Calculate \(1 - \frac{h}{R}\):
\[ 1 - 1.5625 \times 10^{-4} = 0.99984375 \]Step 5: Square this value:
\[ (0.99984375)^2 \approx 0.9996875 \]Step 6: Calculate \(g_h\):
\[ g_h = 9.8 \times 0.9996875 = 9.796 \, \text{m/s}^2 \]Answer: The acceleration due to gravity at 1000 m above Earth's surface is approximately \(9.80 \, \text{m/s}^2\), slightly less than at the surface.
Step 1: Calculate weight on Earth:
\[ W_{\text{Earth}} = m \times g_{\text{Earth}} = 70 \times 9.8 = 686 \, \text{N} \]Step 2: Calculate weight on Moon:
\[ W_{\text{Moon}} = m \times g_{\text{Moon}} = 70 \times 1.62 = 113.4 \, \text{N} \]Answer: The person weighs 686 N on Earth and 113.4 N on the Moon.
Step 1: Identify given values:
Step 2: Use formula for gravity at depth \(d\):
\[ g_d = g \left(1 - \frac{d}{R}\right) \]Step 3: Calculate \(\frac{d}{R}\):
\[ \frac{500}{6.4 \times 10^6} = 7.8125 \times 10^{-5} \]Step 4: Calculate \(g_d\):
\[ g_d = 9.8 \times (1 - 7.8125 \times 10^{-5}) = 9.8 \times 0.999921875 = 9.799 \, \text{m/s}^2 \]Answer: The acceleration due to gravity 500 m below the surface is approximately \(9.799 \, \text{m/s}^2\), a very slight decrease.
Step 1: Use the formula for weight:
\[ W = mg \]Step 2: Rearrange to find mass:
\[ m = \frac{W}{g} = \frac{98}{9.8} = 10 \, \text{kg} \]Answer: The mass of the object is 10 kg.
When to use: While solving problems involving forces between two bodies.
When to use: To simplify calculations in entrance exam problems.
When to use: Especially in mixed-unit problems.
When to use: When height or depth is much smaller than Earth's radius.
When to use: In theoretical and descriptive questions.
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