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.

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:

$F=G\frac{Mm}{R^2}$

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:

$F=10N=G\frac{Mm}{R^2}$   (1)

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

$G\frac{M}{R^2}=g$   (2)

Then:

$10N=g.m_{1}=g.(1kg)$   (3)

Finding $g$:

$g=\frac{10N}{1kg}=10m/s^2$  (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 $R+h=3R/2$:

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

Substituting (4) in (7):

$F_{h}=g\frac{4m_{2}}{9}$   (8)

Knowing the mass of the research satellite is 200kg:

$F_{h}=10m/s^2\frac{(4)(200kg)}{9}$  (9)

Finally:

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