A rocket is launched vertically from Earth's surface with a velocity of 3.4 km/s. How high does it go:
From Earth's centre?
From Earth's surface?
how can this be done without any time or force mentioned ??
To determine the height the rocket reaches from Earth's center or surface, we need to use the concept of escape velocity.
The escape velocity is the minimum velocity needed for an object to escape the gravitational pull of a particular planet or body. For Earth, the escape velocity is approximately 11.2 km/s.
Since the given velocity of the rocket (3.4 km/s) is less than the escape velocity, it will not escape Earth's gravitational pull. Therefore, the rocket will eventually fall back to Earth.
To calculate the maximum height the rocket reaches from Earth's center or surface, we can use the principles of projectile motion.
1. From Earth's Center:
In this case, we assume the rocket starts at the Earth's center and moves upwards. Since there is no resistance or other forces acting on it (other than gravity), the height it reaches will be exactly halfway between its initial position and the point at which it falls back to Earth.
Using this information, we can calculate the height using the following formula:
Height = (Radius of Earth)/2
The radius of Earth is approximately 6,371 km. Therefore, the height the rocket reaches from Earth's center would be:
Height = 6,371 km / 2 = 3,185.5 km
2. From Earth's Surface:
In this case, we assume the rocket starts from the surface of the Earth and moves upwards. Again, the height it reaches will be the maximum height before it falls back to Earth.
To calculate the maximum height from Earth's surface, we need to determine the total energy of the rocket, which consists of both its kinetic energy and potential energy at its highest point.
Using the conservation of energy, we can equate the initial kinetic energy of the rocket to its potential energy at the highest point:
(1/2)mv^2 = mgh
Where:
m = mass of the rocket (which we assume to be constant)
v = velocity of the rocket (3.4 km/s)
g = acceleration due to gravity (approximately 9.8 m/s^2, which we convert to km/s^2)
Simplifying the equation:
(1/2)(3.4 km/s)^2 = (9.8 km/s^2)h
Solving for h (the maximum height):
h = (3.4 km/s)^2 / (2 * 9.8 km/s^2) = 1.16 km
Therefore, the rocket reaches a maximum height of approximately 1.16 km from Earth's surface.
In conclusion, for a rocket launched vertically from Earth's surface with a velocity of 3.4 km/s, the maximum height it reaches from Earth's center is approximately 3,185.5 km, and from Earth's surface is approximately 1.16 km.
To calculate the height reached by the rocket, we can make certain assumptions and use basic principles of projectile motion.
Assuming no air resistance, no external forces acting on the rocket after launch, and that gravity is the only force acting on the rocket.
From Earth's Surface:
We can calculate the maximum height reached by the rocket using the concept of potential and kinetic energy. At maximum height, all of the initial kinetic energy is converted into potential energy.
1. First, convert the initial velocity from km/s to m/s.
Initial velocity (v) = 3.4 km/s = 3,400 m/s
2. Using the conservation of energy, equate the initial kinetic energy (KE) to the potential energy (PE) at the maximum height.
KE = PE
(1/2)mv^2 = mgh
Where m is the mass of the rocket, g is the acceleration due to gravity, and h is the maximum height reached.
3. The mass of the rocket cancels out:
(1/2)v^2 = gh
4. Rearrange the equation to solve for h:
h = (1/2)v^2 / g
5. Finally, substitute the values and calculate the height:
g = 9.8 m/s^2 (approximate acceleration due to gravity on Earth)
h = (1/2)(3,400)^2 / 9.8
h ≈ 580,204 meters or 580.2 km
From Earth's Center:
To calculate the height from the Earth's center, we need to subtract the radius of the Earth from the height calculated from the Earth's surface.
Earth's radius (r) = 6,371 km
Height from Earth's center = Height from Earth's surface - Earth's radius
Height from Earth's center ≈ 580.2 km - 6,371 km
Height from Earth's center ≈ -5,790.8 km
Note: The negative sign indicates that the rocket cannot go below the Earth's surface. Therefore, the highest point reached by the rocket from the Earth's center is approximately 5,790.8 km.
vₒ= 3.4 km/s= 3400 m/s.
The problem may be solved by several methods.
The first method (kinematics)
The height from the Earth's surface
h =vₒ•t -g•t²/2,
v=vₒ-g•t.
At the top point v=0, => vₒ=g•t, t = vₒ/g,
Substitute it in the equation for h and obtain
h=vₒ²/2•g =(3400 )²/2•9.8 ≈ 5.9•10^5 m. =590 km.
The second method (the law of conservation of energy)
KE = ΔPE.
m•v²/2 = m•g•h,
h = vₒ²/2•g =….
The Earth’s radius is R ≈ 6378 km.
The distance from Earth's centre is
H = 6378 + 590 = 6968 km.