Problem 21 - Master Resource Book in JEE Main Physics

Question

An object weighs \(10 \mathrm{~N}\) at the north-pole of the earth. In a geostationary satellite distant \(7 R\) from the centre of earth (of radius \(R\) ), what will be its true weight? (a) \(3 \mathrm{~N}\) (b) \(5 \mathrm{~N}\) (c) \(2 \mathrm{~N}\) (d) \(0.2 \mathrm{~N}\)

Step-by-Step Solution

Answer

(d) 0.2 N

Step 1: Understanding the Problem

We are given that an object weighs 10 N at the north pole of the Earth. We are asked to find its weight in a geostationary satellite 7 Earth radii away from the center of the Earth.

Step 2: Recall the Gravitational Weight Formula

The weight of an object due to gravity is governed by the formula: \[ W = mg \]where \( W \) is the weight, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity. However, at a distance from the Earth's surface, gravity changes according to the formula:\[ g' = \frac{GM}{r^2} \]where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the distance from the center of the Earth.

Step 3: Apply the Weight Change Formula

Since the weight is inversely proportional to the square of the distance from the center of the Earth, the weight at any distance \( r \) is given by:\[ W' = W \times \left( \frac{R}{r} \right)^2 \]Here, \( R \) is the Earth’s radius, and \( r = 7R \). Therefore:\[ W' = 10 \times \left( \frac{R}{7R} \right)^2 \]

Step 4: Simplify and Calculate the Weight

Substitute the known value of \( r = 7R \) into the equation:\[ W' = 10 \times \left( \frac{1}{7} \right)^2 \]\[ W' = 10 \times \frac{1}{49} \]\[ W' = \frac{10}{49} \approx 0.204 \mathrm{~N} \]

Step 5: Compare and Choose the Right Option

From the above calculation, the weight of the object in the geostationary satellite is approximately 0.204 N, which matches option (d) 0.2 N.

Key Concepts

Weight Calculations

Weight is the force exerted by gravity on an object. It's calculated using the formula: \( W = mg \). Here, \( W \) represents the weight, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity, which is approximately \( 9.81 \text{ m/s}^2 \) at Earth's surface.

• The mass of an object remains constant regardless of its location.
• However, the weight can change based on the gravitational force acting on it.
• On Earth's surface, weight is often referred to in units of newtons (N).

At different distances from the center of the Earth, the impact of gravity changes, altering the weight calculation. For example, in a geostationary satellite, the gravitational force is weaker compared to that at the poles. This difference is crucial for understanding why an object's weight decreases as it moves further from Earth, as seen in this exercise.

Geostationary Satellite

A geostationary satellite orbits the Earth at a fixed position relative to the planet's surface. It matches Earth's rotation, appearing stationary in the sky.

• These satellites orbit approximately 35,786 kilometers above Earth's equator.
• Their main purpose is for telecommunications, weather forecasting, and broadcasting.
• Such satellites maintain a constant distance from the Earth, which is crucial for consistent data transmission.

When calculating the gravitational effects on an object within a geostationary satellite, it's important to consider that the satellite is at a much greater distance—often expressed in terms of Earth's radii—from the Earth's center than locations on its surface. As in the exercise, the object was found to be 7 radii away, significantly reducing the gravitational pull, and thus its weight.

Inverse Square Law

The Inverse Square Law describes how a physical quantity or intensity decreases with increasing distance from its source. For gravitational force, it means the force weakens with the square of the distance:
\[ g' = \frac{GM}{r^2} \]

• \( G \) is the gravitational constant, \( M \) is Earth's mass.
• \( r \) is the distance from the object's center to Earth's center.
• This formula explains why objects in orbit experience significantly less gravitational force.

In the given exercise, the weight decrease is computed using the inverse square law: \[ W' = W \times \left( \frac{R}{r} \right)^2 \]. By plugging in \( r = 7R \), it is clear that weight diminishes as the square of the distance increases. Therefore, when moving away from Earth, the gravitational influence decreases rapidly, explaining the much-reduced weight of 0.2 N for the object at the geostationary satellite's location.