Home
Class 12
PHYSICS
The total energy of a satellite is...

The total energy of a satellite is

A

always positive

B

always negative

C

always zero

D

positive or negative depending upon radius of orbit

Text Solution

AI Generated Solution

The correct Answer is:
To determine the total energy of a satellite in orbit, we can follow these steps: ### Step 1: Understand the Components of Total Energy The total energy (E) of a satellite in orbit is the sum of its kinetic energy (K) and potential energy (U). ### Step 2: Calculate the Kinetic Energy The kinetic energy of the satellite can be expressed as: \[ K = \frac{1}{2} mv^2 \] where \( m \) is the mass of the satellite and \( v \) is its orbital velocity. ### Step 3: Determine the Orbital Velocity The orbital velocity \( v \) of the satellite can be derived from the gravitational force acting on it. The formula for orbital velocity is: \[ v = \sqrt{\frac{GM}{r}} \] where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the distance from the center of the Earth to the satellite. ### Step 4: Substitute the Velocity into Kinetic Energy Substituting the expression for \( v \) into the kinetic energy formula gives: \[ K = \frac{1}{2} m \left(\sqrt{\frac{GM}{r}}\right)^2 \] \[ K = \frac{1}{2} m \frac{GM}{r} \] \[ K = \frac{GMm}{2r} \] ### Step 5: Calculate the Potential Energy The gravitational potential energy (U) of the satellite is given by: \[ U = -\frac{GMm}{r} \] ### Step 6: Find the Total Energy Now, we can find the total energy (E) by adding kinetic and potential energy: \[ E = K + U \] \[ E = \frac{GMm}{2r} - \frac{GMm}{r} \] \[ E = \frac{GMm}{2r} - \frac{2GMm}{2r} \] \[ E = -\frac{GMm}{2r} \] ### Step 7: Analyze the Result The total energy \( E \) is negative, which indicates that the satellite is in a bound state in its orbit. ### Conclusion Thus, the total energy of a satellite in orbit is always negative. ### Final Answer The total energy of a satellite is always negative. ---
Promotional Banner

Similar Questions

Explore conceptually related problems

A satellite of mass m is orbiting the earth (of radius R ) at a height h from its surface. The total energy of the satellite in terms of g_(0) , the value of acceleration due to gravity at the earth's surface,

A satellite of mass m is orbiting the earth (of radius R ) at a height h from its surface. The total energy of the satellite in terms of g_(0) , the value of acceleration due to gravity at the earth's surface,

At what height from the surface of the earth, the total energy of satellite is equal to its potential energy at a height 2R from the surface of the earth ( R =radius of earth)

If potential energy of a satellite is -2 MJ ,then the binding energy of satellite is .

A: When a satellite is orbiting then no energy is required to keep moving n its orbit . R : The total mechanical energy of a satellite is conserved .

Mark the correct statement (i) Escape velocity does not depend on mass of body (ii) If total energy of a satellite becomes positive , it escape from earth . (iii) Orbit of geostationary orbit is called parking orbit

The total energy of an electron is 3.555 MeV, then its Kinetic energy is

Assertion : The binding energy of a satellite does not depend upon the mass of the satellite. Reason : Binding energy is the negative value of total energy of satellite.

A satellite of mass m is just placed over the surface of earth. In this position mechanical energy of satellite is E_(1) . Now it starts orbiting round the earth in a circular path at height h = radius of earth. In this position, kinetic energy potential energy and total mechanical energy of satellite are K_(2) , U_(2) and E_(2) respectively. Then

The correct graph representing the variation of total energy (E_(t)) , kinetic energy (E_(k)) and potential energy (U) of a satellite with its distance form the centre of earth is