Question

A satellite of mass m is orbiting the earth of r radius at a height h from its surface.The total energy of the satellite in terms of g0 , the value of acceleration due to gravity at the earth's surface is,

Answer 1

Answer:

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The total energy of the satellite in terms of $g_{0}$, acceleration due to gravity at the earth's surface = $\huge \frac{ - m g_{0} {r}^{2} }{2(r + h)}$

Explanation:

Hint:-

• We know that the Total energy of a satellite = sum of its potential energy and kinetic energy
• The potential energy of a system due to Earth's gravitational pull is always -ve.
• The acceleration due to gravity is the constant acceleration acting on a body above the Earth's surface.

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Total energy of satellite is:-

E = $\large\frac{ - GmMm}{2(r + h)}$ ....(1)

Here, E is the total energy of the satellite

G is the gravitational constant

M is the mass of the Earth

m is the mass of the satellite

R is the radius of the Earth

h is the height of satellite from the Earth’s surface

Now,

Acceleration due to gravity, $g_{0}$ = $\large\frac{GM}{ {r}^{2} }$ ....(2)

From (2)

$g_{0} {r}^{2}$ = GM

$\therefore$ E = $\large\frac{ - g_{0} {r}^{2} m}{2(r + h)}$

Answer 2

Answer:

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