krit.club logo

Gravitation - Energy of an Orbiting Satellite

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

A satellite revolving around a planet of mass MM and radius RR at a distance rr from the center possesses both Kinetic Energy (KK) and Potential Energy (UU).

•

The Potential Energy (UU) of a satellite of mass mm at distance rr is given by U=−GMmrU = -\frac{GMm}{r}. The negative sign indicates that the satellite is in a bound state within the gravitational field.

•

The Kinetic Energy (KK) is derived from the orbital velocity vo=GMrv_o = \sqrt{\frac{GM}{r}}, resulting in K=12mvo2=GMm2rK = \frac{1}{2} m v_o^2 = \frac{GMm}{2r}.

•

The Total Mechanical Energy (EE) is the sum of Kinetic and Potential energies: E=K+U=−GMm2rE = K + U = -\frac{GMm}{2r}.

•

Relationship between energies: E=−K=U2E = -K = \frac{U}{2}.

•

Binding Energy is the minimum energy required to remove the satellite from its orbit to infinity. It is equal to the magnitude of the Total Energy: B.E.=−E=GMm2rB.E. = -E = \frac{GMm}{2r}.

•

If the satellite is at a height hh from the surface of the Earth, then r=R+hr = R + h.

📐Formulae

U=−GMmrU = -\frac{GMm}{r}

K=GMm2rK = \frac{GMm}{2r}

E=K+U=GMm2r−GMmr=−GMm2rE = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}

vo=GMrv_o = \sqrt{\frac{GM}{r}}

B.E.=GMm2rB.E. = \frac{GMm}{2r}

💡Examples

Problem 1:

Calculate the energy required to move a satellite of mass mm from a circular orbit of radius 2R2R to a radius 3R3R, where RR is the radius of the Earth and MM is the mass of the Earth.

Solution:

The total energy of a satellite in an orbit of radius rr is E=−GMm2rE = -\frac{GMm}{2r}. Initial energy at r1=2Rr_1 = 2R: E1=−GMm2(2R)=−GMm4RE_1 = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R} Final energy at r2=3Rr_2 = 3R: E2=−GMm2(3R)=−GMm6RE_2 = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R} The energy required (Work done) is ΔE=E2−E1\Delta E = E_2 - E_1: ΔE=−GMm6R−(−GMm4R)\Delta E = -\frac{GMm}{6R} - \left( -\frac{GMm}{4R} \right) ΔE=GMmR(14−16)\Delta E = \frac{GMm}{R} \left( \frac{1}{4} - \frac{1}{6} \right) ΔE=GMmR(3−212)=GMm12R\Delta E = \frac{GMm}{R} \left( \frac{3 - 2}{12} \right) = \frac{GMm}{12R} Using GM=gR2GM = gR^2, the energy is mgR12\frac{mgR}{12}.

Explanation:

Energy required is the difference between the final total energy and the initial total energy. Since the final orbit is further away, the total energy becomes less negative (increases), necessitating an input of energy.

Problem 2:

A satellite of mass 2000 kg2000\text{ kg} is in an orbit. If its Potential Energy is −10×109 J-10 \times 10^9\text{ J}, calculate its Kinetic Energy and Total Energy.

Solution:

We know the relationship between energies for an orbiting satellite:

  1. Kinetic Energy K=−12UK = -\frac{1}{2} U
  2. Total Energy E=12UE = \frac{1}{2} U Given U=−10×109 JU = -10 \times 10^9\text{ J}: K=−12(−10×109)=5×109 JK = -\frac{1}{2} (-10 \times 10^9) = 5 \times 10^9\text{ J} E=12(−10×109)=−5×109 JE = \frac{1}{2} (-10 \times 10^9) = -5 \times 10^9\text{ J} To check, E=K+UE = K + U: 5×109−10×109−5×109\begin{array}{r} 5 \times 10^9 \\ - 10 \times 10^9 \\ \hline - 5 \times 10^9 \end{array}

Explanation:

The Kinetic Energy of a satellite is always half the magnitude of its Potential Energy and positive, while the Total Energy is half the Potential Energy and remains negative.