Question

At a certain distance from the center of the Earth, a 0.4-kg object has a weight of 2.0 N. (a) Find this distance. (b) If the object is released at this location and allowed to fall toward the Earth, what is its initial acceleration

Answer 1

Answer:

a) The distance of the object from the center of the Earth is 8.92x10⁶ m.

b) The initial acceleration of the object is 5 m/s².

Explanation:

a) The distance can be found using the equation of gravitational force:

$F = \frac{GMm}{r^{2}}$

Where:

G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²

M: is the Earth's mass =  5.97x10²⁴ kg  

m: is the object's mass = 0.4 kg

F: is the force or the weight = 2.0 N    

r: is the distance =?

The distance is:

$r = \sqrt{\frac{GMm}{F}} = \sqrt{\frac{6.67 \cdot 10^{-11} Nm^{2}/kg^{2}*5.97 \cdot 10^{24} kg*0.4 kg}{2.0 N}} = 8.92 \cdot 10^{6} m$      

Hence, the distance of the object from the center of the Earth is 8.92x10⁶ m.

         

b) The initial acceleration of the object can be calculated knowing the weight:              

$W = ma$                                                  

Where:            

W: is the weight = 2 N

a: is the initial acceleration =?          

$a = \frac{W}{m} = \frac{2 N}{0.4 kg} = 5 m/s^{2}$

Therefore, the initial acceleration of the object is 5 m/s².

           

I hope it helps you!