Answer:
Centre of mass is the point at which the distribution of mass is equal in all directions, and does not depend on gravitational field. Centre of gravity is the point at which the distribution of weight is equal in all directions, and does depend on gravitational field.
Answer:
High speed optical communication technology
To be able to communicate from the space to the earth and from earth to space is one of the most essential features required during space exploration.
Explanation:
Space exploration involves going into the space, beyond the earth's atmosphere. Landing on other planets and studying their details, going into deeper space beyond the planets to discover new cosmic events or structures is all a part of space exploration.
The key to analyse the studies and observations is being able to communicate the data collected, photos taken etc to the launch centers or space centers on earth. The space centers on earth should also be able to communicate with the persons or the satellites in space.
This is made possible using the optical communication technology which involves the use of optical fibers, lasers etc, since high speeds are more efficient during communication
Answer: 1010.92 m/s
Explanation:
According to Newton's law of universal gravitation:
(1)
Where:
is the gravitational force between Earth and Moon
is the Gravitational Constant
is the mass of the Earth
is the mass of the Moon
is the distance between the Earth and Moon
Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to :
(2)
Where 
is the centripetal acceleration given by:
(3)
Being 
the orbital velocity of the moon
Making (1)=(2):
(4)
Simplifying:
(5)
Making (5)=(3):
(6)
Finding :
(7)
(8)
Finally:
Answer:
K = -½U
Explanation:
From Newton's law of gravitation, the formula for gravitational potential energy is;
U = -GMm/R
Where,
G is gravitational constant
M and m are the two masses exerting the forces
R is the distance between the two objects
Now, in the question, we are given that kinetic energy is;
K = GMm/2R
Re-rranging, we have;
K = ½(GMm/R)
Comparing the equation of kinetic energy to that of potential energy, we can derive that gravitational kinetic energy can be expressed in terms of potential energy as;
K = -½U
