An object is projected from a satellite orbiting 200 km above the surface of the earth. The object has a speed relative to the Earth of 2.85×104 m/s. What is its speed when it is very far from the earth?
find the GPE of the satellite at 200km . Add that to the KE at 200km. Then, assume it is all KE, find v.
GPE= GMe*m/d where d is distance from center of earth.
v=sqrt(2(-GM/d)+1/2(vi^2)) based on conservation of energy (gravitational potential energy and kinetic energy).
To determine the speed of the object when it is very far from the Earth, we need to consider the principle of conservation of mechanical energy. When the object is near the Earth's surface, it possesses both kinetic energy and gravitational potential energy. As it moves higher, the potential energy increases while the kinetic energy decreases.
Given:
Height of satellite above the surface of the Earth (h) = 200 km = 200,000 m
Speed of the object relative to the Earth (v) = 2.85 × 10^4 m/s
To find the speed of the object when it is very far from the Earth, we assume that the object has reached a point where its gravitational potential energy is negligible. At that point, its total mechanical energy is equal to its kinetic energy.
The equation for the total mechanical energy (E) is given by the sum of kinetic energy (K) and gravitational potential energy (U):
E = K + U
When the object is at the surface of the Earth, its kinetic energy is given by:
K = (1/2)mv^2
where m is the mass of the object.
The gravitational potential energy is given by:
U = mgh
where g is the acceleration due to gravity and h is the height above the surface.
Since the mass of the object cancels out, we can rewrite the equation as:
E = (1/2)v^2 + gh
Now we can find the total mechanical energy when the object is near the Earth's surface:
E1 = (1/2)(2.85 × 10^4)^2 + (9.8)(0)
As the object moves far away from the Earth, its height (h) becomes very large. At this point, we assume that the gravitational potential energy is negligible compared to the kinetic energy, so the equation becomes:
E2 ≈ (1/2)v^2 + (9.8)(∞)
Since (∞) is undefined, we can say that E2 ≈ (1/2)v^2
Since energy is conserved, E1 = E2:
(1/2)(2.85 × 10^4)^2 + (9.8)(0) ≈ (1/2)v^2
Now we can solve for v:
(1/2)(2.85 × 10^4)^2 ≈ (1/2)v^2
(2.85 × 10^4)^2 ≈ v^2
v ≈ ±2.85 × 10^4 m/s
Therefore, the speed of the object when it is very far from the Earth is approximately ±2.85 × 10^4 m/s.