a satellite of mass 100 kg is rotating around earth in a circular orbit if radius 3r.what extra energy should be given to this satellite if it us to be lifted into an orbit of radius 4r?

To determine the extra energy required to lift a satellite from an orbit of radius 3r to a higher orbit of radius 4r, we need to consider the gravitational potential energy and the kinetic energy associated with each orbit. The total mechanical energy of a satellite in orbit is the sum of its kinetic and potential energies, and we can calculate the difference between the two orbits to find the extra energy needed.

Understanding Orbital Energy

The gravitational potential energy (U) of a satellite in orbit is given by the formula:

U = -\frac{GMm}{r}

Where:

• G is the gravitational constant (approximately 6.674 × 10^-11 N(m/kg)^2)
• M is the mass of the Earth (about 5.972 × 10^24 kg)
• m is the mass of the satellite (100 kg in this case)
• r is the radius of the orbit

Calculating Potential Energy at Different Orbits

First, let's calculate the gravitational potential energy at both orbits:

For the orbit at radius 3r:

U1 = -\frac{GMm}{3r}

For the orbit at radius 4r:

U2 = -\frac{GMm}{4r}

Finding the Change in Potential Energy

The change in gravitational potential energy (ΔU) when moving from the orbit at 3r to 4r is:

ΔU = U2 - U1

Substituting the expressions for U1 and U2:

ΔU = -\frac{GMm}{4r} - \left(-\frac{GMm}{3r}\right)

This simplifies to:

ΔU = -\frac{GMm}{4r} + \frac{GMm}{3r}

To combine these fractions, we need a common denominator, which is 12r:

ΔU = \frac{GMm}{3r} - \frac{GMm}{4r} = \frac{4GMm - 3GMm}{12r} = \frac{GMm}{12r}

Calculating Kinetic Energy

The kinetic energy (K) of a satellite in orbit is given by:

K = \frac{1}{2}mv^2

For a circular orbit, the orbital speed (v) can be derived from the gravitational force providing the necessary centripetal force:

F_gravity = F_centripetal

\frac{GMm}{r^2} = \frac{mv^2}{r}

From this, we can solve for v:

v = \sqrt{\frac{GM}{r}}

Now, we can calculate the kinetic energy for both orbits:

For radius 3r:

K1 = \frac{1}{2}m\left(\sqrt{\frac{GM}{3r}}\right)^2 = \frac{1}{2}m\frac{GM}{3r} = \frac{GMm}{6r}

For radius 4r:

K2 = \frac{1}{2}m\left(\sqrt{\frac{GM}{4r}}\right)^2 = \frac{1}{2}m\frac{GM}{4r} = \frac{GMm}{8r}

Calculating the Total Energy for Each Orbit

The total energy (E) of the satellite in each orbit is the sum of its kinetic and potential energies:

For radius 3r:

E1 = K1 + U1 = \frac{GMm}{6r} - \frac{GMm}{3r} = \frac{GMm}{6r} - \frac{2GMm}{6r} = -\frac{GMm}{6r}

For radius 4r:

E2 = K2 + U2 = \frac{GMm}{8r} - \frac{GMm}{4r} = \frac{GMm}{8r} - \frac{2GMm}{8r} = -\frac{GMm}{8r}

Finding the Extra Energy Required

The extra energy (ΔE) needed to lift the satellite from the orbit at 3r to 4r is:

ΔE = E2 - E1

Substituting the values we calculated:

ΔE = \left(-\frac{GMm}{8r}\right) - \left(-\frac{GMm}{6r}\right)

This simplifies to:

ΔE = -\frac{GMm}{8r} + \frac{GMm}{6r}

Finding a common denominator (24r):

ΔE = \frac{-3GMm + 4GMm}{24r} = \frac{GMm}{24r}

Final Calculation

Now, substituting the values:

m = 100 kg

G = 6.674 × 10^-11 N(m/kg)^2

M = 5.972 × 10^24 kg

r = radius of Earth (approximately 6.371 × 10^6 m)

Plugging these values into the energy equation:

ΔE = \frac{(6.674 × 10^{-11})(5.972 × 10^{24})(100)}{24(6.371 × 10^{6})}

Calculating this will give you the extra energy required to lift the satellite from an orbit of radius 3r to 4r. This approach illustrates the principles of gravitational potential energy and kinetic energy in orbital mechanics, providing a comprehensive understanding of