Two satellites A and B of the same mass
are orbiting Earth in concentric orbits. The
distance of satellite B from Earth’s center is
twice that of satellite A.
What is the ratio of the tangential speed of
B to that of A?
The tangential speed of an object in circular motion is given by the formula:
v = √(G*M/r)
Where v is the tangential speed, G is the gravitational constant, M is the mass of the Earth, and r is the distance between the object and the Earth's center.
Let's assume the distance between satellite A and the Earth's center is r, so the distance between satellite B and the Earth's center is 2r.
The mass of the satellites doesn't affect the tangential speed, so we can ignore it in this problem.
The tangential speed of satellite A is:
vA = √(G*M/r)
The tangential speed of satellite B is:
vB = √(G*M/2r)
To find the ratio of the tangential speeds, we can divide vB by vA:
vB/vA = (√(G*M/2r))/(√(G*M/r))
Using the property of square roots (√(a/b) = √a / √b), we can simplify:
vB/vA = (√(G*M/2r))/(√(G*M/r))
vB/vA = (√(G*M* r/2r))/(√(G*M/r))
Canceling out r, we have:
vB/vA = (√(G*M/2))/(√(G*M))
vB/vA = √(1/2)
Therefore, the ratio of the tangential speed of satellite B to that of satellite A is √(1/2).
To find the ratio of the tangential speed of satellite B to that of satellite A, we need to consider the concept of centripetal force. The centripetal force acting on an object in circular motion is provided by the gravitational force between the object and the center it's orbiting.
The gravitational force depends on the mass of the object and the distance between the centers of the two objects. Since both satellites have the same mass, the centripetal force acting on them will be the same.
The centripetal force can be expressed as:
Fc = (mv²) / r
Where Fc is the centripetal force, m is the mass of the object, v is the tangential speed, and r is the distance from the center of the orbit.
For satellite A:
Fa = (mA * vA²) / rA
For satellite B:
Fb = (mB * vB²) / rB
Since Fa = Fb, we can equate the two equations:
(mA * vA²) / rA = (mB * vB²) / rB
Given that the distance of satellite B from Earth's center is twice that of satellite A (rB = 2rA), we can substitute this into the equation:
(mA * vA²) / rA = (mB * vB²) / (2rA)
Next, we can simplify the equation by canceling out the masses:
vA² / rA = vB² / (2rA)
Now, let's solve for the ratio of tangential speeds:
vB / vA = √[(vA² / rA) / (vB² / (2rA))]
Simplifying further, we get:
vB / vA = √(2)
Therefore, the ratio of the tangential speed of satellite B to that of satellite A is √2.
Vc = sqrt(µ/r)
V(A)^2 = µ/r(A)
V(B)^2 = µ/r(B) = µ/2r(A)
V(B)^2/(2/V(A)^2 = [µ/2r(A)]/[µr(A)]
.................= 2
V(B)/V(A) = sqrt2