In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.00×1016 and a radius of 8.50 .
What is the speed of a satellite orbiting 4.70km above the surface?
What is the escape speed from the asteroid?
You need to provide units for your quantities. I will have to assume your 8.50 is km and your 10^16 is kg.
4.70 km above the surface is R = 13.2 km (or 1.32*10^4 m) from the center
The orbital speed satisfies
G M/R^2 = V^2/R
where M is the asteroid mass and you know what G is. Solve for V.
V = sqrt(GM/R)
For the escape velocity Ve from the asteroid FROM THE SURFACE, where R' = 8.50 *10^3 m,
Ve^2/2 = GM/R'
Ve = sqrt(2 G M/R')
Note that a different radius, R', must be used.
Why did the satellite go to therapy? Because it had some serious orbit issues!
To calculate the speed of the satellite orbiting 4.70 km above the surface, we can use the formula for the orbital speed:
V = √(G * M / R)
Where:
V = orbital speed
G = gravitational constant (approximated to 6.67 × 10^-11 m^3/kg/s^2)
M = mass of the asteroid (1.00 × 10^16 kg)
R = radius of the asteroid + altitude (8.50 km + 4.70 km = 13.20 km = 13,200 m)
Plugging in the values, we get:
V = √((6.67 × 10^-11) * (1.00 × 10^16) / (13,200))
V = √(6.67 × 10^5)
V ≈ 816.5 m/s
Therefore, the speed of the satellite orbiting 4.70 km above the surface is approximately 816.5 m/s.
Now, let's calculate the escape speed from the asteroid! Hold on tight, things are about to get speedy!
The escape speed can be found using the formula:
Ve = √(2 * G * M / R)
Where:
Ve = escape speed
Plugging in the values, we get:
Ve = √(2 * (6.67 × 10^-11) * (1.00 × 10^16) / 13,200)
Ve = √(1.35 × 10^6)
Ve ≈ 1162.4 m/s
So the escape speed from the asteroid is approximately 1162.4 m/s. Whoosh, that's fast! Don't forget your anti-gravity suit!
To calculate the speed of a satellite orbiting 4.70 km above the surface of the asteroid, we will use the formula for the orbital speed:
v = √(G * M / r)
where:
v is the orbital speed,
G is the gravitational constant (6.67430 × 10^-11 N m^2 / kg^2),
M is the mass of the asteroid, and
r is the distance of the satellite from the center of the asteroid, which is the sum of the asteroid's radius and the altitude of the satellite.
Given:
Mass of the asteroid (M) = 1.00 × 10^16 kg
Radius of the asteroid (R) = 8.50 km (8.50 × 10^3 m)
Altitude of the satellite (h) = 4.70 km (4.70 × 10^3 m)
Step 1: Convert the values to SI units.
Radius of the asteroid (R) = 8.50 × 10^3 m
Altitude of the satellite (h) = 4.70 × 10^3 m
Step 2: Calculate the distance of the satellite from the center of the asteroid.
r = R + h
r = 8.50 × 10^3 m + 4.70 × 10^3 m
r = 13.20 × 10^3 m
Step 3: Substitute the values into the formula for orbital speed.
v = √(G * M / r)
v = √((6.67430 × 10^-11 N m^2 / kg^2) * (1.00 × 10^16 kg) / (13.20 × 10^3 m))
Calculating this expression will give you the speed of the satellite orbiting above the surface of the asteroid.
To calculate the escape speed from the asteroid, we will use the formula:
v = √(2 * G * M / r)
where:
v is the escape speed,
G is the gravitational constant,
M is the mass of the asteroid, and
r is the distance from the center of the asteroid at which the object is located.
Step 4: Substitute the values into the formula for escape speed.
v = √(2 * G * M / r)
v = √(2 * (6.67430 × 10^-11 N m^2 / kg^2) * (1.00 × 10^16 kg) / (13.20 × 10^3 m))
Calculating this expression will give you the escape speed from the asteroid.
To find the speed of a satellite orbiting a spherical body, we can use the formula for the orbital speed:
v = √(G * M / r)
where v is the orbital speed, G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2), M is the mass of the asteroid, and r is the distance between the center of the asteroid and the satellite.
In this case, the mass of the asteroid is given as 1.00 × 10^16 kg, and the radius of the asteroid is given as 8.50 km (or 8.50 × 10^3 m). The distance between the satellite and the surface of the asteroid is 4.70 km (or 4.70 × 10^3 m).
Let's calculate the orbital speed:
Step 1: Convert all units to SI units (meters, kilograms, and seconds):
Mass of the asteroid (M) = 1.00 × 10^16 kg
Radius of the asteroid (r) = 8.50 × 10^3 m
Distance from satellite to surface (r') = 4.70 × 10^3 m
Step 2: Plug the values into the formula:
v = √(G * M / r)
= √[(6.67430 × 10^-11 m^3 kg^-1 s^-2) * (1.00 × 10^16 kg) / (8.50 × 10^3 m)]
≈ √(6.67430 × 10^5 m^2 s^-2 / 8.50 × 10^3 m)
Step 3: Calculate the orbital speed:
v ≈ √(6.67430 × 10^5 / 8.50 × 10^3) m/s
≈ √(78,525) m/s
≈ 280.20 m/s (rounded to two decimal places)
Therefore, the speed of the satellite orbiting 4.70 km above the surface of the asteroid is approximately 280.20 m/s.
Now, let's find the escape speed from the asteroid.
The escape speed is the minimum speed an object needs to escape from the gravitational pull of a celestial body. It is given by the formula:
v_escape = √(2 * G * M / r)
where v_escape is the escape speed, G is the gravitational constant, M is the mass of the asteroid, and r is the radius of the asteroid.
Using the same values for the mass and radius of the asteroid as before, let's calculate the escape speed:
v_escape = √(2 * G * M / r)
= √[(2 * 6.67430 × 10^-11 m^3 kg^-1 s^-2) * (1.00 × 10^16 kg) / (8.50 × 10^3 m)]
Simplifying the equation:
v_escape = √(2 * 6.67430 × 10^5 / 8.50 × 10^3) m/s
= √(156,058) m/s
≈ 394.98 m/s (rounded to two decimal places)
Therefore, the escape speed from the asteroid is approximately 394.98 m/s.