with what velocity must a body be thrown from earth surface so that it may reach a height 4Re the earth surface. given Re=6400km g=9.8m/s
now
F = m g
work done against gravity = integral F*dr or m g dr
or
m * integral g dr
m 9.8 Re^2 integral dr/r^2
from Re to 4 Re
integral dx/x^2 = -1/x
so here
1/6,400,000 - 1/25,600,000
= 1.17 * 10^-7
so work done in Joules -
m 9.8 Re^2 * 1.17* 10^-7
= m (47,040,000) Joules
THAT = (1/2)mv^2
so
v = 9,700 meters/second
To find the velocity required for the body to reach a height of 4Re (where Re is the radius of the Earth), we can use the principle of conservation of mechanical energy.
The potential energy of the body at height h is given by the formula PE = mgh, where m is the mass of the body, g is the acceleration due to gravity, and h is the height.
The kinetic energy of the body when it is thrown is given by the formula KE = (1/2)mv^2, where m is the mass of the body and v is the velocity.
At the initial point of throwing, both the potential energy and kinetic energy are zero.
At the highest point, the potential energy is maximum (mgh) and the kinetic energy is zero.
According to the principle of conservation of mechanical energy, the total mechanical energy (sum of potential and kinetic energy) remains constant throughout the motion.
Therefore, we can equate the initial mechanical energy to the final mechanical energy:
0 + 0 = mgh + (1/2)mv^2
Since h = 4Re and g = 9.8 m/s^2, we can substitute these values into the equation:
0 + 0 = mg(4Re) + (1/2)mv^2
Since m is the mass of the body, it cancels out from both sides of the equation:
0 + 0 = g(4Re) + (1/2)v^2
Simplifying further:
0 = 4gRe + (1/2)v^2
Rearranging the equation:
(1/2)v^2 = -4gRe
v^2 = -8gRe
Taking the square root of both sides:
v = √(-8gRe)
Given g = 9.8 m/s^2 and Re = 6400 km = 6400000 m, we substitute these values into the equation:
v = √(-8 * 9.8 * 6400000)
Calculating this expression:
v ≈ √(-5017600000)
Since the result is imaginary (square root of a negative number), it means that it is not possible for a body to be thrown from the Earth's surface with a velocity such that it reaches a height of 4Re (where Re is the radius of the Earth).
To find the velocity with which a body must be thrown from the Earth's surface to reach a height of 4 times the Earth's radius (4Re), we can use the principles of projectile motion and energy conservation.
Step 1: Determine the initial potential energy and final potential energy.
The initial potential energy (U1) is given by:
U1 = mgh1
Where:
m = mass of the object (assumed to be constant)
g = acceleration due to gravity (9.8 m/s^2)
h1 = initial height (Earth's surface)
The final potential energy (U2) is given by:
U2 = mgh2
Where:
h2 = final height (4Re)
Step 2: Equate the initial potential energy to the final potential energy.
U1 = U2
mgh1 = mgh2
Step 3: Substitute the known values.
We know that h1 is zero (since we are starting from the Earth's surface) and h2 is 4 times the Earth's radius (4Re = 4 * 6400 km).
m * 9.8 * 0 = m * 9.8 * (4 * 6400 km)
Step 4: Convert the units.
Since we have different units for acceleration due to gravity (m/s^2) and height (km), let's convert km to meters.
4 * 6400 km = 4 * 6400 * 1000 m = 25,600,000 m
Step 5: Cancel out the mass (m) on both sides.
We can cancel out the mass (m) since it appears on both sides of the equation.
9.8 * 0 = 9.8 * 25,600,000
Step 6: Solve for the velocity (v).
Divide both sides of the equation by 9.8 to solve for the velocity (v).
v = 25,600,000 m / 9.8 s^2
Calculating this, we get:
v ≈ 2,612,244.9 m/s
Therefore, the body must be thrown from the Earth's surface with a velocity of approximately 2,612,244.9 m/s to reach a height of 4 times the Earth's radius.
Although g may be about 9.8 m/s^2 at the surface of earth, it is not 9.8 at 4 Re and therefore you can not assume it is constant
g = 9.8 (Re^2/r^2)