need help with the following question. if earth has a mass of 6.0*10^24 kg and a radius 6.4*10^6 m:
a) find the change in gravitational potential energy of a meteorite of mass 150 kg as it moves from an infinite distance to the earth's surface.
b) with what speed does the meteorite strike the earth if it starts from rest?
thanks. i kinda get the answer for part (a) but (b) is mostly a problem.
To solve part (b), we need to consider the conservation of energy. The change in gravitational potential energy of the meteorite is converted into its kinetic energy as it falls towards the Earth's surface.
Step 1: Calculate the change in gravitational potential energy (ΔPE) of the meteorite:
The gravitational potential energy (PE) is given by the formula:
PE = -GMm/r
where G is the gravitational constant (6.67 × 10^-11 N m^2/kg^2), M is the mass of the Earth, m is the mass of the meteorite, and r is the distance from the center of the Earth to the meteorite.
Given:
Mass of the Earth (M) = 6.0 × 10^24 kg
Radius of the Earth (r) = 6.4 × 10^6 m
Mass of the meteorite (m) = 150 kg
First, we need to calculate the potential energy at the initial distance (infinity) from the surface of the Earth. At infinity, the potential energy is zero, so:
PE_initial = 0
Next, we calculate the potential energy at the surface of the Earth:
PE_final = -GMm/r
Step 2: Calculate the change in potential energy (ΔPE):
ΔPE = PE_final - PE_initial
= -GMm/r
Substituting the given values:
ΔPE = -6.67 × 10^-11 N m^2/kg^2 × 6.0 × 10^24 kg × 150 kg / (6.4 × 10^6 m)
= -5.3475 × 10^6 J (to four significant figures)
Therefore, the change in gravitational potential energy of the meteorite as it moves from an infinite distance to the Earth's surface is -5.3475 × 10^6 J.
Step 3: Calculate the kinetic energy (KE) of the meteorite at the Earth's surface:
ΔKE = -ΔPE (by the conservation of energy)
= 5.3475 × 10^6 J
Step 4: Apply the formula for kinetic energy:
KE = 1/2 mv^2
where m is the mass of the meteorite and v is its velocity.
Rearranging the formula for velocity:
v = √(2KE/m)
Substituting the given values:
v = √(2 × 5.3475 × 10^6 J / 150 kg)
= √(71.3 × 10^3 m^2/s^2 / kg)
= √(71.3 × 10^3 m^2 s^-2 kg^-1)
= √(71.3 × 10^3 N kg^-1)
= √(71.3 × 10^3) m/s
≈ 267.4 m/s
Therefore, the meteorite strikes the Earth with a velocity of approximately 267.4 m/s.
To find the change in gravitational potential energy of a meteorite as it moves from an infinite distance to the Earth's surface, we can use the formula:
ΔPE = -GMm/r
Where:
- ΔPE is the change in gravitational potential energy
- G is the gravitational constant (which is approximately 6.67 * 10^-11 N.m^2/kg^2)
- M is the mass of the Earth
- m is the mass of the meteorite
- r is the distance from the center of the Earth to the location where the meteorite is moving (in this case, the Earth's surface)
a) Calculating the change in gravitational potential energy:
ΔPE = -GMm/r
Given:
- Mass of Earth (M) = 6.0 * 10^24 kg
- Radius of Earth (r) = 6.4 * 10^6 m
- Mass of the meteorite (m) = 150 kg
Substituting the given values:
ΔPE = - (6.67 * 10^-11 N.m^2/kg^2) * (6.0 * 10^24 kg) * (150 kg) / (6.4 * 10^6 m)
Now, let's calculate this:
ΔPE = - 2.25 * 10^8 J
The change in gravitational potential energy of the meteorite as it moves from an infinite distance to the Earth's surface is approximately -2.25 * 10^8 J.
b) To find the speed at which the meteorite strikes the Earth, we can use the principle of conservation of energy. We can equate the change in potential energy (ΔPE) to the kinetic energy (KE) gained by the meteorite.
ΔPE = KE
Kinetic energy can be calculated using the formula:
KE = 0.5 * mv^2
Where:
- KE is the kinetic energy gained by the meteorite
- m is the mass of the meteorite
- v is the velocity of the meteorite
Setting the change in potential energy equal to the kinetic energy:
ΔPE = KE
-2.25 * 10^8 J = 0.5 * (150 kg) * v^2
Solving for v:
v^2 = (2 * -2.25 * 10^8 J) / (150 kg)
v^2 = -3.0 * 10^6 m^2/s^2
Since velocity cannot be negative, we can take the positive square root:
v = √(-3.0 * 10^6 m^2/s^2)
v ≈ 1732 m/s
Therefore, the meteorite will strike the Earth with a speed of approximately 1732 m/s if it starts from rest.
I hope that helps! Let me know if you have any further questions.