The Hubble Space Telescope orbits at an altitude of 600 km above the Earth's surface. Assuming it is in a stable circular orbit, what is its velocity? For reference, the Earth's radius is 6400 km and Earth's mass is 6 x 10^24 kg.
The formula to solve for circular velocity is v(circ) = the square root of (GM/r)
The answer is 7500 m/s but I'm not sure how to get to this answer from this equation
What you want to do is add the altitude to the earth's radius to get the orbital radius. In your case, you'd add 600 km to 6400 km to get 7000 km. Use 7000 km as your r value and plug it into the formula. When you're trying to find r, measure the distance from the center of the earth (or planet I guess) to the object.
Vcirc= square root of (GM/r)
Here's a similar example:
G = 6.67 × 10^−11 m3 / (kg s2)
MEarth = 5.98 × 10^24 kg
rEarth = 6.4 × 10^6 m
altitude of roughly 6 × 10^5 meters
Add altitude of object and radius of earth to get orbital radius:
r= 7 x 10^6
Use this r value in the formula: Vcirc= square root of (GM/r)
In my example, you'll get ~7579 m/s
To find the velocity of the Hubble Space Telescope in a stable circular orbit, we can use the formula for circular velocity:
v(circ) = √(GM/r)
Where:
v(circ) is the circular velocity,
G is the gravitational constant (6.67 x 10^-11 N m^2/kg^2),
M is the mass of the Earth (6 x 10^24 kg),
and r is the orbital radius of the Hubble Space Telescope (600 km above the Earth's surface).
First, let's convert the given orbital altitude to meters by multiplying by 1000:
r = 600 km x 1000 = 600,000 meters.
Now we can substitute the values into the formula:
v(circ) = √(6.67 x 10^-11 N m^2/kg^2 x 6 x 10^24 kg / 600,000 m)
Calculating this equation step by step:
1. Multiply the mass of the Earth by the gravitational constant:
6.67 x 10^-11 N m^2/kg^2 x 6 x 10^24 kg = 40.02 x 10^13 N m.
2. Divide the result by the orbital radius:
40.02 x 10^13 N m / 600,000 m = 66.7 x 10^7 N.
3. Take the square root of the division result:
√(66.7 x 10^7 N) ≈ 8162.2 m/s.
Rounding this value to the nearest whole number, the velocity of the Hubble Space Telescope in a stable circular orbit is approximately 8162 m/s or simply 8162 m/s (since it's not realistic to have decimal places for velocity).
Thus, the answer is approximately 8162 m/s, which is slightly different from the given answer of 7500 m/s. Please double-check the given answer to ensure its accuracy.
To find the velocity of the Hubble Space Telescope in a stable circular orbit, you can use the formula for circular velocity, which is v(circ) = √(GM/r).
Here's how to calculate it step by step:
1. First, let's determine the values for the variables in the formula:
- G is the universal gravitational constant, which is approximately 6.67430 x 10^(-11) m^3kg^(-1)s^(-2).
- M represents the mass of Earth, which is 6 x 10^24 kg.
- r is the radius of the orbit, which is the sum of the Earth's radius and the altitude of the telescope: r = 6400 km + 600 km = 7000 km = 7 x 10^6 m.
2. Now that we have the values, let's substitute them into the formula:
v(circ) = √(GM/r)
= √((6.67430 x 10^(-11) m^3kg^(-1)s^(-2)) * (6 x 10^24 kg) / (7 x 10^6 m))
3. Simplify the equation inside the square root:
v(circ) = √((4.00458 x 10^14 m^3kg^(-1)s^(-2)) / (7 x 10^6 m))
4. Divide the numbers:
v(circ) = √(5.72083 x 10^7 m^2s^(-2))
= 7,560 m/s (rounded to three significant figures)
Therefore, the velocity of the Hubble Space Telescope in a stable circular orbit is approximately 7,560 m/s.
r=6.4E6+.6E6 m
Does that do it for you?