Question

) A 500 kg satellite is in a circular orbit with an orbital radius of 12,000 km. (a) Draw a force digram of the satellite and calculate the gravitational force and the acceleration for the satellite. Show directions of both the force and the acceleration. (b) Calculate the Earth’s gravitational field at the location of the satellite. (c) Calculate the satellite’s velocity and orbital period. (d) Find the kinetic energy and gravitational potential energy of the satellite. (e) Calculate the energy required to move the satellite into a higher circular orbit of radius 24,000 km.Radius of Earth = 6,380 km Mass of Earth = 6.0 x 1024 kg

Answer 1

Given:

The mass of the satellite, m=500 kg

The orbital radius of the satellite, r=12,000 km= 12,000×10³ m

The higher orbital radius of the satellite, h=24000 km=12000×10³ m

The mass of the earth, M=6.0×10²⁴ kg

The radius of the earth R=6380 km=6380×10³ m

To find:

a) Force diagram, gravitational force, and acceleration of the satellite.

b) Gravitational field strength of the earth at the location of the satellite.

c) Satellite's velocity and orbital period,

d) Kinetic energy and the gravitational potential energy of the satellite.

e) The energy required to move the satellite into a higher orbital.

Step-by-step explanation:

The force diagram:

Referring to the diagram, F is the gravitational force, a is the acceleration of the satellite and v is the velocity of the satellite.

The gravitational force is given by,

`F=(GMm)/((R+r)^2)`

Where G is the gravitational constant.

On substituting the known values,

`\begin{gathered} F=(6.67*10^(-11)*6*10^(24)*500)/((6380*10^3+12000*10^3)^2) \\ =592.3\text{ N} \end{gathered}`

The force is also given by,

`F=mg_r`

Where g_r is the acceleration of the satellite or the gravitational field strength of the earth at the location of the satellite.

On substituting the known values,

`\begin{gathered} 592.3=500* g_r \\ \Rightarrow g_r=(592.3)/(500) \\ =1.18\text{ m/s}^2 \end{gathered}`

b)

As calculated in part a, the gravitational field strength of the earth at the position of the satellite is g_r=1.18 m/s²

c)

The orbital velocity of the satellite is given by,

`v=\sqrt{(GM)/((R+r))}`

On substituting the known values,

`\begin{gathered} v=\sqrt{(6.67*10^(-11)*6*10^(24))/((6380*10^3+12000*10^3))} \\ =7266.14\text{ m/s} \end{gathered}`

The orbital period of the satellite is given by,

`T=(2\pi(R+r))/(v)`

On substituting the known values,

`\begin{gathered} T=(2\pi(6380*10^3+12000*10^3))/(7266.14) \\ =15893.6\text{ s} \end{gathered}`

d) The kinetic energy of the satellite is given by,

`KE=(1)/(2)mv^2`

On substituting the known values,

`\begin{gathered} KE=(1)/(2)*500*7266.14^2 \\ =1.32*10^(10)\text{ J} \end{gathered}`

The potential energy of the satellite is given by,

`PE=mg_r(R+r)`

On substituting the known values,

`\begin{gathered} PE=500*1.18*(6380*10^3+12000*10^3) \\ =1.08*10^(10)\text{ J} \end{gathered}`

e)

The energy needed to move the satellite to a higher orbit is given by the difference in the total energy of the satellite in these two orbits.

The orbital velocity of the satellite in the higher orbit is given by,

`v_h=\sqrt{(GM)/((R+h))}`

On substituting the known values,

`\begin{gathered} v_h=\sqrt{(6.67*10^(-11)*6*10^(24))/((6,380*10^3+24,000*10^3))} \\ =6751.4\text{ m/s} \end{gathered}`

Thus the kinetic energy of the satellite will be,

`KE_h=(1)/(2)mv_h^2`

On substituting the known values,

`\begin{gathered} KE_h=(1)/(2)*500*6751.4^2 \\ =1.14*10^(10)\text{ J} \end{gathered}`

The gravitational potential energy of the satellite is given by,

`PE_h=(GMm)/((R+h))`

On substituting the known values,

`\begin{gathered} PE_h=(6.67*10^(-11)*6*10^(24)*500)/((6,380*10^3+24,000*10^3)) \\ =6.6*10^9\text{ J} \end{gathered}`

Thus the energy needed is given by,

`E=KE_h+PE_h-KE-PE`

On substituting teh known values,

`\begin{gathered} E=1.14*10^(10)+6.6*10^9-1.32*10^(10)-1.08*10^(10) \\ =-6*10^9\text{ J} \end{gathered}`

Final answer:

(a) The gravitational force is 592.3 N

The acceleration is 1.18 m/s²

(b) The gravitational field at the location of the satellite is 1.18 m/s²

(c) The orbital velocity is 7266.14 m/s and the period is 15893.6 s

(d) The kinetic energy is 1.32×10¹⁰ J and the potential energy is 1.08×10¹⁰ J

(e) The energy required is -6×10⁹ J