Question

On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m and may be treated as uniform spherical objects.
Find the magnitude of the gravitational force that each part exerts on the other out in space, far from any other objects.

Answer 1

Answer:

$F = 1.489*10^{-7} N$

Explanation: Weight of space probes on earth is given by:$W= m*g$

W= weight of the object( in N)

m= mass of the object (in kg)

g=acceleration due to gravity(9.81 $\frac{m}{s^{2} }$)

Therefore,

$m_{1} = \frac{14500}{9.81}$

$m_{1} = 1478.08 kg$

Similarly,

$m_{2} = \frac{4800}{9.81}$

$m_{2} = 489.29 kg$

Now, considering these two parts as uniform spherical objects

Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.

Force between these two objects is given by:

$F = \frac{Gm_{1} m_{2}}{R^{2} }$

G= gravitational constant ($6.67 * 10^{-11} m^{3} kg^{-1} s^{-2}$)

$m_{1} , m_{2}$= masses of the object

R= distance between their centres (in m)(18 m)

Substituiting all these values into the above formula

$F = 1.489*10^{-7} N$

This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.