Please help On a fictitious planet, you weigh a satellite and find it to be 500 N. The satellite is then put into circularorbit about the planet a distance 12 km above the surface of the planet, where it takes 90 minutes tocomplete an orbit. If the radius of the planetis 1.2x10 m:i) What is the mass of the satellite?) If the satellite was moved to an orbit 9 km above the surface of the planet, does its kinetic energyincrease, decrease, or stay the same?) Does its gravitational potential energy increase, decrease, or stay the same?iv) What is the total energy of the satellite now (in the new orbit]?617..pdf199+

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On a fictitious planet, you weigh a satellite and find it to be 500 N. The satellite is then put into circular
orbit about the planet a distance 12 km above the surface of the planet, where it takes 90 minutes to
complete an orbit. If the radius of the planetis 1.2x10 m:
i) What is the mass of the satellite?
) If the satellite was moved to an orbit 9 km above the surface of the planet, does its kinetic energy
increase, decrease, or stay the same?
) Does its gravitational potential energy increase, decrease, or stay the same?
iv) What is the total energy of the satellite now (in the new orbit]?
617..pdf
199+

Expert Solution

Step 1

If $M$ is the mass of the planet, $R$ is the radius of the planet and $h$ is the height above which the satellite is moving then the time period of revolution of the satellite is

$T = \frac{C i r c u m f e r e n c e o f o r b i t}{O r b i t a l v e l o c i t y} = \frac{2 π (R + h)}{v_{0}}$

The orbital velocity of the satellite is given by

$v_{0} = R \sqrt{\frac{g}{R + h}}$

Step 2

i) Given the weight of the satellite on the planet $W = 500 N$

the height above the ground where the satellite is moving $h = 12 k m = 12 \times 10^{3} m$

The time period of the satellite $T = 90 m i n = 90 \times 60 s = 5400 s$

The radius of the planet $R = 1.2 \times 10^{6}$

Therefore from the expression of the time period

$T = \frac{2 π (R + h)}{v_{0}}$

The orbital velocity of the satellite is given by

$v_{0} = \frac{2 π (R + h)}{T} = \frac{2 π \times (1.2 \times 10^{6} + 12 \times 10^{3})}{5400} = 1410.226 m / s$

If $g$ is the acceleration due to the gravity on the planet, then,

$v_{0} = R \sqrt{\frac{g}{R + h}}$

This gives

$1410.226 = (1.2 \times 10^{6}) \sqrt{\frac{g}{(1.2 \times 10^{6} + 12 \times 10^{3})}} \Rightarrow g = (1.2 \times 10^{6} + 12 \times 10^{3}) \times {(\frac{1410.226}{1.2 \times 10^{6}})}^{2} = 1.674 m / s^{2}$

Therefore the mass of the satellite

$m = \frac{W e i g h t}{A c c e l e r a t i o n d u e t o g r a v i t y} = \frac{W}{g} = \frac{500}{1.674} k g = 298.69 k g$

ii) As the orbital velocity of the satellite is dependent on the height then the kinetic energy of the satellite will also change with the height of the orbit. The orbital velocity is inversely proportional to the height. Therefore as the height decreases the orbital velocity will increase leading to an increase in the kinetic energy of the satellite.

Orbital velocity when it is at a height of $12 k m$ from the surface of the planet $v_{0 i} = 1410.226 m / s$

Mass of the satellite $m = 298.69 k g$

The kinetic energy of the satellite

$K E_{i} = \frac{1}{2} m v_{0 i}^{2} = \frac{1}{2} \times 298.69 \times 1410 . 226^{2} J = 2.97 \times 10^{8} J$

Orbital velocity of the satellite at a height $9 k m$ from the surface of the planet

$v_{0 f} = R \sqrt{\frac{g}{R + h}} = (1.2 \times 10^{6}) \times \sqrt{\frac{1.674}{1.2 \times 10^{6} + 9 \times 10^{3}}} m / s = 1412.03 m / s$

Thus the Kinetic energy of the satellite

$K E_{f} = \frac{1}{2} m v_{0 f}^{2} = \frac{1}{2} \times 298.69 \times 1412 . 03^{2} J = 2.9776 \times 10^{8} J$

iii) If $v_{0}$ is the orbital velocity then it is also expressed as

$v_{0} = \sqrt{\frac{G M}{R + h}}$

$M$ is the mass of the planet and $G$ is the universal gravitational constant. The potential energy at a height h is given by

$U = - \frac{G M m}{R + h} = - m v_{0}^{2}$

At height $h = 12 k m$ the gravitational potential energy is

$U_{i} = - m v_{0 i}^{2} = - (298.69) \times 1410 . 226^{2} = - 5.94 \times 10^{8} J$

At height $h = 9 k m$ , the gravitational potential energy is

$U_{f} = - m v_{0 f}^{2} = - (298.69) \times 1412 . 03^{2} = - 5.955 \times 10^{8} J$

The potential energy has increased once the satellite is moved to a lower orbit.

iv) The total energy of the satellite in the new orbit

$E = K E_{f} + U_{f} = (2.977 \times 10^{8}) - (5.95 \times 10^{8}) = - 2.973 \times 10^{8} J$

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