Eris, the largest dwarf planet known in the Solar System, has a radius R = 1200 km and an acceleration due to gravity on its surface of magnitude g = 0.77 m/s2.
(a) Use these numbers to calculate the escape speed from the surface of Eris.
(b) If an object is fired directly upward from the surface of Eris with half of this escape speed, to what maximum height above the surface will the object rise? (Assume that Eris has no atmosphere and negligible rotation.)
I know the speed is 1360 m/s but how do you solve question b? The answer is 400 km. Thank you in advance.
A)
Escape vel. = Ue = sqrt(2GM/R) ...(1)
also g = GM/R^2 So GM/R = gR .....(2)
From(1) & (2)
Ue = sqrt(2gR) = sqrt(2*0.77*1200,000)
= 1360 m/s
B)
u = Ue/2 = (1/2)*sqrt(2GM/R)
Now KEi+PEi = PEf+KEf (conservation of mechanical energy)
or (1/2)mu^2 - GMm/R = -GMm/(R+h)+0
or GM/4R - GM/R = -GM/(R+h)
or 3/4R = 1/(R+h)
or h = R/3 = 400 Km
I'm having trouble following this. How do you go from:
GM/4R - GM/R = -GM/(R + h) to 3/4R = 1/(R+h) to
h=R/3?
Well, it seems like Eris is quite a small planet, which makes me wonder if it might be feeling a bit inadequate compared to the larger ones. But hey, size doesn't matter, right? Let's move on to the questions.
(a) To calculate the escape speed, we can use the formula v = √(2gR). Plugging in the values, we have v = √(2 * 0.77 m/s² * 1200 km) = √(2 * 0.77 m/s² * 1200000 m) = √(1848000 m²/s²) = 1360 m/s. So, you are correct that the escape speed is 1360 m/s. Now, let's move on to part b.
(b) If an object is fired directly upward from the surface of Eris with half of the escape speed, we can call this half speed v/2 = 1360 m/s / 2 = 680 m/s. We want to find the maximum height above the surface that this object will reach.
To calculate this, we can use the formula for the maximum height of projectile motion, which is given by h = (v² / (2g)).
Plugging in the values, we have h = (680 m/s)² / (2 * 0.77 m/s²) = (462400 m²/s²) / (1.54 m/s²) = 300000 m = 300 km.
So, it looks like my initial calculations came up a little short, but not by much. The object will rise to a maximum height of 300 km above the surface of Eris. Seems like it's reaching for the stars, but maybe it should aim a bit higher and go for that 400 km mark!
To solve question (b), we can use the principle of conservation of mechanical energy.
The total mechanical energy of the object at its maximum height will be equal to the sum of its kinetic energy and potential energy. At its maximum height, the object will momentarily come to rest, so its kinetic energy will be zero.
The kinetic energy of the object is given by the equation:
Kinetic Energy = (1/2)mv^2
where m is the mass of the object and v is its velocity.
Since the object is fired upward with half of the escape speed, its initial velocity is (1/2)(1360 m/s) = 680 m/s.
We are given the object's mass, but since we are only interested in the object's maximum height, we can ignore the mass in our calculations.
Therefore, the initial kinetic energy of the object is:
Initial Kinetic Energy = (1/2)mv^2 = (1/2)(680 m/s)^2 = 231,200 J
At its maximum height, the object has no kinetic energy, so the total mechanical energy is equal to the potential energy:
Potential Energy = Total Mechanical Energy = 231,200 J
The potential energy of the object is given by the equation:
Potential Energy = mgh
where h is the height above the surface.
Rearranging the equation, we get:
h = Potential Energy / (mg)
Substituting the known values, we have:
h = 231,200 J / (m)(g)
h = 231,200 J / (m)(0.77 m/s^2)
Given that the radius of Eris is 1200 km, we can convert it to meters by multiplying by 1000:
R = 1200 km = 1200 x 1000 m = 1,200,000 m
Using the equation for the acceleration due to gravity on the surface of a spherical body, we have:
g = (GM) / (R^2)
Solving for M (mass of Eris), we have:
M = (g)(R^2) / G
Substituting the known values, we have:
M = (0.77 m/s^2)(1,200,000 m)^2 / (6.67 x 10^-11 N m^2/kg^2)
Calculating the mass of Eris, we find:
M = 1.39 x 10^22 kg
Substituting the values for m and g, we have:
h = 231,200 J / (1.39 x 10^22 kg)(0.77 m/s^2)
Simplifying the equation, we find:
h ≈ 400,000 m = 400 km
Therefore, the object will rise to a maximum height of 400 km above the surface of Eris before falling back down.
To solve part (b) of the question, we can use the concept of conservation of energy.
When the object is fired directly upward with half of the escape speed, its initial kinetic energy is given by (1/2)mv^2, where m is the mass of the object and v is the initial velocity.
At the maximum height, the object comes to a stop and all its initial kinetic energy is converted into gravitational potential energy. The gravitational potential energy of an object at a height h above the surface of Eris is given by mgh, where g is the acceleration due to gravity on Eris.
Since the object comes to a stop at its maximum height, its final velocity will be zero. Therefore, we can equate the initial kinetic energy to the gravitational potential energy:
(1/2)mv^2 = mgh
Since we want to find the maximum height above the surface, we can set h as the variable we wish to solve for. We can cancel out the mass 'm' from both sides of the equation:
(1/2)v^2 = gh
Substituting the given values, we have:
(1/2)(1360/2)^2 = (0.77)(h)
Simplifying, we get:
1360^2/8 = 0.77h
Solving for h, we get:
h ≈ 400 km
Therefore, the maximum height above the surface that the object will rise to is approximately 400 km.