Newton's three forces, normal, tension and friction, are present in a surprising number of physical situations
Newton's Laws, that describe the relationship between an obejct and the forces acting upon it, apply in almost every physical situation, from quantum mechanics to electricity.
The correct answer is:
Newton’s laws can explain the forces that occur between objects every day
Answer:
1750 Joules.
Explanation:
Work done = force * distance
= 7 * 250
= 1750 Joules.
To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.
Planet gravitational force



Distance between planet and star

Gravitational force is

Applying the new distance,


Replacing with the previous force,

Replacing our values


Therefore the magnitude of the force on the star due to the planet is 
Answer:

Explanation:
To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.
By conservation of energy we know that,

Where,

Replacing


Our values are given by,





Replacing at the equation,


Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was 