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Question: A research satellite of mass \(200\) kg circles the Earth in an orbit radius \(\dfrac{{3{R_E}}}{2}\)...

A research satellite of mass 200200 kg circles the Earth in an orbit radius 3RE2\dfrac{{3{R_E}}}{2} , where RE{R_E} is the radius of the Earth. Assuming the gravitational pull on a mass of 11 kg on the Earth’s surface to be 10N10\,N , the pull on the satellite will be:
A) 890N890\,N
B) 889N889\,N
C) 885N885\,N
D) 892N892\,N

Explanation

Solution

Gravitational pull on a mass of 11 kg on the Earth’s surface is given as 10N10\,N . Use the formula of force due to the gravitational pull of the Earth. Then divide the force of gravitational pull on mass of 11 kg with the force between Earth and satellite to get the required answer.

Complete step by step solution:
We need to find the force acting on the satellite due to the gravitational pull of the Earth. We are given the value of force on a mass of 11 kg due to gravitational pull of the Earth.
The force FF acting on a body of mass mm which is at a distance of RE{R_E} from the centre of Earth is given as:
F=GMEmRE2F = \dfrac{{G{M_E}m}}{{{R_E}^2}}
Here, ME{M_E} is the mass of Earth.
Now, for a body having mass of 11 kg, this force will be:
F=GME×1RE2=GMERE2F = \dfrac{{G{M_E} \times 1}}{{{R_E}^2}} = \dfrac{{G{M_E}}}{{{R_E}^2}}
But this force is given to be of 10N10\,N .
F=GMERE2=10N\therefore F = \dfrac{{G{M_E}}}{{{R_E}^2}} = 10\,N --equation 11
The mass of the research satellite is m=200m = 200 kg
the distance of the object from the centre of the Earth is 3RE2\dfrac{{3{R_E}}}{2}
Therefore, the gravitational pull on the research satellite will be:
F=GME×200(32RE)2F = \dfrac{{G{M_E} \times 200}}{{{{\left( {\dfrac{3}{2}{R_E}} \right)}^2}}}
F=8009×GMERE2\Rightarrow F = \dfrac{{800}}{9} \times \dfrac{{G{M_E}}}{{{R_E}^2}}
But from equation 11 , we have GMERE2=10N\dfrac{{G{M_E}}}{{{R_E}^2}} = 10\,N , thus:
F=8009×10\Rightarrow F = \dfrac{{800}}{9} \times 10
F=888.8N=889N\therefore F = 888.8\,N = 889\,N
This is the pull on the satellite.

Thus, option C is the correct option.

Note: The given options are very close to the final answer, be careful while calculating the values. We were directly given the value of some constants indirectly so be careful to not put unnecessary values such as mass of Earth and the radius of radius. Instead make use of the given values this will save time and there will be less chances of calculation mistakes.