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\(\Lambda\) satellite of mass \(m\) is in orbit round the carth at a distance of \(2.5 R\) above its surface. If the gravitational field strength at the earth's surface is \(g .\) Calculate the centripetal force acting on the satellite (a) \(\frac{m g}{2}\) (b) \(\frac{4 m g}{25}\) (c) \(\frac{4 m g}{49}\) (d) \(\frac{m g}{25}\)

Short Answer

Expert verified
The centripetal force acting on the satellite is \(\frac{4 mg}{49}\).

Step by step solution

01

Understand the problem

The satellite is orbiting the Earth at a distance of \(2.5 R\) above its surface. We need to calculate the centripetal force acting on it given the gravitational field strength \(g\) at Earth's surface.
02

Calculate the orbital radius

The distance from the center of the Earth (radius \(R\)) to the satellite is the sum of Earth's radius \(R\) and the distance above the surface \(2.5 R\). Thus, the orbital radius \(r\) is \(r = R + 2.5R = 3.5R\).
03

Use the formula for gravitational force

The gravitational force \(F_g\) acting on the satellite is given by the formula \(F_g = \frac{GMm}{r^2}\), where \(G\) is the gravitational constant, \(M\) is Earth's mass, and \(r\) is the distance from the Earth's center.
04

Relate gravitational force to surface gravity

At Earth's surface, the gravitational force can also be written as \(mg = \frac{GMm}{R^2}\). We can use this relation to replace \(\frac{GM}{R^2}\) with \(g\) in the previous formula.
05

Substitute and simplify the gravitational force

Replace \(\frac{GM}{r^2}\) using the relation from Earth's surface: \(F_g = \frac{GMm}{r^2} = mg \cdot \frac{R^2}{r^2}\). Substitute \(r = 3.5R\) into this equation: \(F_g = mg \cdot \frac{R^2}{(3.5R)^2}\).
06

Calculate the expression

Simplify \(\frac{R^2}{(3.5R)^2}\) to \(\frac{1}{(3.5)^2}\). Calculating \(3.5^2\) gives \(12.25\). So, \(F_g = mg \times \frac{1}{12.25}\).
07

Matching with given options

Convert \(\frac{1}{12.25}\) to a fraction: \(\frac{1}{12.25} = \frac{1}{\left(\frac{49}{4}\right)} = \frac{4}{49}\). Therefore, \(F_g = \frac{4mg}{49}\). This matches option (c).

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Key Concepts

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

Gravitational Field Strength
Gravitational field strength, often denoted by the symbol \( g \), is a measure of the force exerted by gravity on a unit mass at a point in space. At the surface of the Earth, it measures approximately \( 9.8 \ m/s^2 \), indicating that a 1 kg object experiences a gravitational force of 9.8 N towards the Earth's center.

This concept is crucial because it helps us calculate the gravitational force acting on any object in proximity to the Earth. The formula for gravitational field strength is \( g = \frac{F}{m} \), where \( F \) is the force due to gravity and \( m \) is the mass.

The gravitational field strength decreases with increasing distance from the surface of the Earth. This is because the gravitational force is inversely proportional to the square of the distance from the center of the Earth, described by 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 Earth's center. As the satellite orbits at a distance \( 3.5R \) from the Earth's center, the gravitational field it experiences is considerably weaker than that at Earth's surface.
Orbital Motion
Orbital motion refers to the motion of an object as it travels around a larger body due to the influence of gravity. For a satellite in orbit around the Earth, the gravitational force acts as a centripetal force, keeping the satellite in a stable path around the planet.

This centripetal force is crucial for maintaining the satellite's orbit and is calculated using the expression \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass of the satellite, \( v \) is its orbital velocity, and \( r \) is the orbital radius.

As derived from the problem, when a satellite orbits at \( 3.5R \) from Earth's center, the centripetal force acting on it due to gravity is given by \( F_g = mg \times \frac{R^2}{(3.5R)^2} \). This position causes the force to significantly decrease compared to when the object is at the Earth's surface.
Earth's Gravitational Field
Earth's gravitational field is the region of space around the planet in which an object experiences a force that pulls it towards the Earth's center. This field is responsible for the phenomena we experience such as objects falling and satellites remaining in orbit.

The strength and effect of Earth's gravitational field change with distance. It diminishes as we move further from the planet. For a satellite orbiting at a distance of \( 2.5R \) above the Earth's surface, this means it is sensing a weaker gravitational pull compared to objects on the surface.

Earth's gravitational field can be quantified using the formula \( g = \frac{GM}{r^2} \). This allows us to compute the necessary conditions for a satellite to stay in orbit at a specific altitude. The decrease in field strength as the distance increases explains the difference in the gravitational force exerted on a satellite compared to that on an object on Earth's surface.

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Most popular questions from this chapter

\(\Lambda\) n object is taken from a point \(P\) to another point \(Q\) in a gravitational field. (a) \Lambdassuming the carth to be spherical, if both \(P\) and \(Q\) lie on carth's surface the work done is zero. (b) If \(P\) is on carth's surface and \(Q\) above it, the work done is minimum when it is taken along the straight line \(P Q\). (c) The work done depends only on the position of \(P\) and \(Q\) and is independent of the path along which the particle is taken. (d) There is no net work done if the object is taken from \(P\) to \(Q\) and then brought back to \(P\), along any path.

In a double star, two stars one of mass \(m_{1}\) and another of mass \(m_{2}\), with a separation \(d\), rotate about their common centre of mass. Find (a) an expression for their time period of revolution, (b) the ratio of their kinetic energies, (c) the ratio of their angular momenta about the centre of mass, and (d) the total angular momentum of the system, (e) the kinetic energy of the system

A man can jump vertically to a height of \(1.5 \mathrm{~m}\) on the carth. Calculate the radius of a planct of the same mean density as that of the carth from whose gravitational ficld he could cscape by jumping. Radius of earth is \(6.41 \times 10^{6} \mathrm{~m}\).

Which of the following are correct? (a) Moon has no atmosphere like carth because the root mean square velocity of all gases is more than their escape velocity from moon's surface. (b) The tangential acceleration of a planet is zero. (c) The atmosphere is held to carh by gravity. (d) nuelcar forec, viscous forec and clectric force are independent of the gravitational cfrect.

A particle of mass \(m\) is subjected to an atuactive central force of magnitude \(k / r^{2}, k\) being a constant. If at the insunt when the paruicle is at an extreme position in its closed path, al a distance \(a\) from the cenure of the forec, its speed is \(\sqrt{\frac{k}{2 m a}}\), find the other extreme position.

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