Question

║⊕QUESTION⊕║
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CLASS 11
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A satellite at a height 'h' from the surface of the earth has the Kinetic energy E. Write the ratio of its kinetic energy to potential energy.

Answer 1

Explanation:

Hello user,

$\huge\bf{\mid{\overline{\underline{Answer:-}}\mid}}$

$\huge{\bold{Given:-}} \\ \\ Satellite \: is \: at \: height = h \\ \\ \huge{\bold{Formula:-}} \\ \\ K. E = \frac{1}{2} \times m_s \times {v}^{2} = \frac{1}{2} \times m \times ({\sqrt{\frac{G \times M_e \times M_s }{R + h}}})^{2} \\ \\ P. E = - \frac{G \times M_e \times M_s }{R + h} \\ \\ Let \: M_e = Mass \: of \: earth \\ \\ m_s = Mass \: of \: satellite \\ \\ Now, \: let's \: find \: answer \: to \: your \: question..... \\ \\ \frac{K.E}{P.E} = \frac{\frac{1}{2} \times m_s \times \frac{G \times M_e }{R+h}}{- \frac{G \times M_e \times m_s }{R+h}} \\ \\ After \: cancelling \: terms \: we'll \: get \implies \\ \\ \boxed{\bold{ \frac{K.E}{P.E} = - \frac{1}{2}}}$

Answer 2

◆【SOLUTION】◆

● Potential energy

Potential energy is somewhat related to our daily life and we come through such millions of example around us. Potential energy is the energy stored in the body when the body move against Conservative field it generally stores in the body.

du = -f.dr

du = -f.drwhere du small potential energy

du = -f.drwhere du small potential energyF is the force

du = -f.drwhere du small potential energyF is the forcedr is the displacement

We can not calculate potential energy as it is based on configuration which is changing continually.

So we set a reference point and calculate the change In your question I chose infinity as the reference and if we come toward earth than potential energy decreases.

◆Formula◆

U = -GMm/r

r is measured from centre of earth

● Potential energy ●

U = -GMm/(R+h)

|U| = GMm/(R+h)....(1)

m is mass of satellite

● Kinetic energy ●

At this height centripetal force balance gravitational force so..

mv²/(R + h) = GMm/(R+h)²

mv² = GMm/(R+h)

mv²/2 = GMm/2(R+h)...... (2)

E(2) = GMm/2(R+h)

This is translational kinetic energy

Divide (2) by (1)

E : U = GMm/2(R+h) : GMm/(R+h)

◆【E : U = 1 : 2】◆

Ratio can't be negative

As you not mentioned that Sattelite revolving or not so this answer is right when Sattelite is not revolving.

If it revolve than it has additional Rotaional kinetic energy. Written as...

E(1) = (1/2)Iw²

where I is the moment of inertia = m(R+h) ²

E(1) = m(R+h)²w²/2

We know that wr = v

So w²(R+h)² = v²

Total kinetic energy = E(1) + E(2)

Total kinetic energy = 1/2mv² + 1/2mv²

Total kinetic energy = mv²

From (2) we get

E (total) = GMm/(R+h).... (3)

Divide (3) by (1)

E : U = 1 : 1