Question

A research satellite of mass 200 kg circles the earth in an 3R orbit of average radiuswhere R is the radius of earth. Assuming the gravitational pull on a mass of 1 kg on the earth's surface to be 10 N, the pull on the satellite will be (a) 880 N (c) 885 N (b) 889 N (d) 892 N

Answer 1

Answer: (b) 889N

The given options do not match with the given data. However, working with an orbit of radius 3R/2 the answer is 889N.

Step-by-step explanation:

According to Newton's law of Gravitation, the force
`F` exerted between two bodies of masses
`M` and
`m` and separated by a distance
`R` is equal to the product of their masses and inversely proportional to the square of the distance:

`F=G(Mm)/(R^2)`

Where
`G`is the gravitational constant,
`M` is the mass of the Earth in this case

.

If we are told the gravitational pull on a mass of
`m_(1)=1 kg`on the earth's surface to be 10 N, this means the force at the surface is:

`F=10N=G(Mm)/(R^2)` (1)

Assuming the mass of the Earth and its radius constant that we will name
`g`, we can say:

`G(M)/(R^2)=g` (2)

Then:

`10N=g.m_(1)=g.(1kg)` (3)

Finding
`g`:

`g=(10N)/(1kg)=10m/s^2` (4)

Now, the gravitational pull at a certain height
`h` over the surface is:

`F_(h)=G\frac{Mm_(2)}{{(R+h)}^2}` (5)

Where
`m_(2)` is the mass of the satellite.

If we are told the radius of the orbit of the research satellite is
`3R/2`, this means
`R+h=3R/2`:

`F_(h)=G\frac{Mm_(2)}{{(3R/2)}^2}=G(M)/(R^2).(4m_(2))/(9)` (7)

Substituting (4) in (7):

`F_(h)=g(4m_(2))/(9)` (8)

Knowing the mass of the research satellite is 200kg:

`F_(h)=10m/s^2((4)(200kg))/(9)` (9)

Finally:

`F_(h)=888.888N\approx 889N` This is the gravitational pull on the satellite with an orbit of radius 3R/2