-GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.
Given
A particle of mass m moving under the influence of a fixed mass's M, gravitational potential energy of formula -GMm/r, where r is the separation between the masses and G is the gravitational constant of the universe.
As the Gravity Potential energy of particle = -GMm/r
Total energy of particle = Kinetic energy + Potential Energy
As we know that
Kinetic energy = 1/2mv²
Also, v is equals to square root of GM/r
v = √GM/r
Put the value of v in the formula of kinetic energy
We get,
Kinetic Energy = GMm/2r
Total Energy = GMm/2r + (-GMm/r)
= GMm/2r - GMm/r
= -GMm/2r
Hence, -GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.
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Answer:
≈ 2.1 R
Explanation:
The moment of inertia of the bodies can be calculated by the equation
I = ∫ r² dm
For bodies with symmetry this tabulated, the moment of inertia of the center of mass
Sphere = 2/5 M R²
Spherical shell = 2/3 M R²
The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass
I = + M D²
Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation
Let's start with the spherical shell, axis is along a diameter
D = 2R
Ic = + M D²
Ic = 2/3 MR² + M (2R)²
Ic = M R² (2/3 + 4)
Ic = 14/3 M R²
The sphere
Is = + M [²
Is = Ic
2/5 MR² + M ² = 14/3 MR²
² = R² (14/3 - 2/5)
= √ (R² (64/15)
= 2,066 R
Answer:
1) Mass that needs to be converted at 100% efficiency is 0.3504 kg
2) Mass that needs to be converted at 30% efficiency is 1.168 kg
Explanation:
By the principle of mass energy equivalence we have
where,
'E' is the energy produced
'm' is the mass consumed
'c' is the velocity of light in free space
Now the energy produced by the reactor in 1 year equals
Thus the mass that is covertred at 100% efficiency is
Part 2)
At 30% efficiency the mass converted equals
The magnitude of the air drag is 784 N
Explanation:
An object falling down reaches the terminal velocity when the magnitude of the air drag acting on it becomes equal to the weight of the object. Mathematically, this condition can be written as:
where
is the magnitude of the air drag
m is the mass of the object
g is the acceleration of gravity
In this problem, we have
m = 80 kg is the mass of the airman
is the acceleration of gravity
Substituting into the formula, we find:
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