The gravitational acceleration at any distance r is given by
where G is the gravitational constant, M the Earth's mass and r is the distance measured from the center of the Earth.
The Earth's radius is 
, so the meteoroid is located at a distance of:
And by substituting this value into the previous formula, we can find the value of g at that altitude:
Given the equation for the Speed of a Satellite
v = SqRt{Gravitational Constant}{Mass of Earth} divided by the radius given in your problem
we have:
(square root whole term on right side)
v = G Me
———
r
so. (6.67x10^-11)(5.97x10^24)
___________________
(8.0x10^6)
v = 7055 m/s (which is reasonable)
so utilize the Kinetic Energy Formula
KE = 1/2mv^2
KE = 1/2(200)(7055)^2
KE = 4.977x10^9 J
Answer:
800 N
Explanation:
By Newton's third law which states that for every action, there is an equal and opposite reaction.
So, as the earth attracts the person towards its center, the person attracts the earth towards itself with the same magnitude of force but in the opposite direction.
Since the person is attracted towards the center of the earth by an 800 N gravitational force, the the earth is attracted toward the person with an 800 N reaction force.
Answer:
38 cm from q1(right)
Explanation:
Given, q1 = 3q2 , r = 60cm = 0.6 m
Let that point be situated at a distance of 'x' m from q1.
Electric field must be same from both sides to be in equilibrium(where EF is 0).
=> k q1/x² = k q2/(0.6 - x)²
=> q1(0.6 - x)² = q2(x)²
=> 3q2(0.6 - x)² = q2(x)²
=> 3(0.6 - x)² = x²
=> √3(0.6 - x) = ± x
=> 0.6√3 = x(1 + √3)
=> 1.03/2.73 = x
≈ 0.38 m = 38 cm = x
