Question

The acceleration of a body due to the attraction of the earth (radius R) at a distance \(2 \mathrm{R}\) from the surface of the earth is \(=\) (g \(=\overline{\text { acceleration due to gravity at the surface of earth })}\) (A) \(\mathrm{g} / 9\) (B) \(\mathrm{g} / 3\) (C) \(\mathrm{g} / 4\) (D) 9

Short answer

The acceleration of a body due to the attraction of the earth at a distance \(2R\) from the surface of the earth is (A) \(g/9\).

Step-by-step solution

Recall gravitational force formula

The formula for gravitational force between two masses (m1 and m2) separated by distance r is given by: \(F = G\frac{m1*m2}{r^2}\) where G is the gravitational constant. In this case, m1 is the mass of the Earth (M) and m2 is the mass of the body (m). The distance r in our case is the sum of Earth's radius and the given distance from the surface, so r = R + 2R = 3R.

Find gravitational acceleration at 2R distance

Divide the gravitational force by the mass of the body (m) to get the acceleration due to gravity (a) at distance 2R from the surface of the Earth: \(a = \frac{F}{m} = \frac{G\frac{M*m}{(3R)^2}}{m} = G\frac{M}{9R^2}\)

Relate the gravitational acceleration at 2R to the Earth's surface

We know that the acceleration due to gravity at the surface of the Earth (g) is given by: \(g = G\frac{M}{R^2}\) Now, we'll divide the acceleration at 2R (a) by the acceleration due to gravity at the surface of the Earth (g): \(\frac{a}{g} = \frac{G\frac{M}{9R^2}}{G\frac{M}{R^2}} = \frac{1}{9}\)

Choose the correct answer

Since the acceleration at 2R distance from the surface of the Earth is g/9, the correct answer is: (A) g/9

Key concepts

Gravitational Force

Gravitational force is a fundamental interaction that attracts two bodies towards each other. It depends on the masses involved and the distance between their centers. The formula for gravitational force (\(F\)) is expressed as:

• \(F = G\frac{m_1*m_2}{r^2}\)

Here\(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the respective masses, and \(r\) is the distance between these masses.
Gravitational force plays a crucial role in maintaining the orbits of planets, moons, and artificial satellites. Despite being a relatively weak force compared to others like electromagnetism, it has an infinite range. Thus, it impacts vast astronomical bodies across the universe.

Distance from Earth's Surface

The term 'distance from Earth's surface' refers to how far an object is situated away from the point where Earth's atmosphere ends and space begins. This distance influences the gravitational pull an object experiences.
For our problem, we are concerned with a distance of \(2R\), where \(R\) is Earth's radius. Linearly adding up, it makes the total distance from the Earth's center \(3R\), including Earth's radius itself.
This is important because the gravitational force decreases with an increase in distance. Mathematically, it follows an inverse square law, meaning the force is inversely proportional to the square of the distance from Earth's center.
As you move away from Earth's surface, the gravitational pull exerted on an object weakens, making the understanding of distance pertinent in gravitational calculations.

Acceleration due to Gravity

Acceleration due to gravity is the rate at which an object will accelerate when it is in free fall under the sole influence of Earth's gravitational pull. On Earth's surface, this value, denoted as \(g\), is approximately \(9.81 \, m/s^2\).
However, as you move further away from the Earth's surface, this acceleration decreases. At \(2R\) distance from the surface, our gravitational acceleration calculation revealed it to be \(g/9\). This occurrence is because the formula:

• \(a = G\frac{M}{r^2}\)

directly ties acceleration to the inverse of the square of the distance \(r\). Therefore, knowing how to calculate acceleration at different distances is key for understanding motion in a gravitational field.
This knowledge is useful in predicting satellite behavior and planning space missions, as it aids in calculating the velocity required to maintain an orbit without the object falling back to Earth.