The given options do not match with the given data. However, working with an orbit of radius 3R/2 the answer is 889N.
Explanation:
According to Newton's law of Gravitation, the force [tex]F[/tex] exerted between two bodies of masses [tex]M[/tex] and [tex]m[/tex] and separated by a distance [tex]R[/tex] is equal to the product of their masses and inversely proportional to the square of the distance:
[tex]F=G\frac{Mm}{R^2}[/tex]
Where [tex]G[/tex]is the gravitational constant, [tex]M[/tex] is the mass of the Earth in this case
.
If we are told the gravitational pull on a mass of [tex]m_{1}=1 kg[/tex]on the earth's surface to be 10 N, this means the force at the surface is:
[tex]F=10N=G\frac{Mm}{R^2}[/tex] (1)
Assuming the mass of the Earth and its radius constant that we will name [tex]g[/tex], we can say:
[tex]G\frac{M}{R^2}=g[/tex] (2)
Then:
[tex]10N=g.m_{1}=g.(1kg)[/tex] (3)
Finding [tex]g[/tex]:
[tex]g=\frac{10N}{1kg}=10m/s^2[/tex] (4)
Now, the gravitational pull at a certain height [tex]h[/tex] over the surface is:
[tex]F_{h}=G\frac{Mm_{2}}{{(R+h)}^2}[/tex] (5)
Where [tex]m_{2}[/tex] is the mass of the satellite.
If we are told the radius of the orbit of the research satellite is [tex]3R/2[/tex], this means [tex]R+h=3R/2[/tex]:
[tex]F_{h}=G\frac{Mm_{2}}{{(3R/2)}^2}=G\frac{M}{R^2}.\frac{4m_{2}}{9}[/tex] (7)
Substituting (4) in (7):
[tex]F_{h}=g\frac{4m_{2}}{9}[/tex] (8)
Knowing the mass of the research satellite is 200kg:
[tex]F_{h}=10m/s^2\frac{(4)(200kg)}{9}[/tex] (9)
Finally:
[tex]F_{h}=888.888N\approx 889N[/tex] This is the gravitational pull on the satellite with an orbit of radius 3R/2
