an asteroid orbits the sun at a constant distance of 4.44e+11 meters, the suns mass is 1.00*10^30kg, what is the orbital speed of the asteroid? Please explain step by step thanks
An asteroid orbits the sun at a constant distance of 4.44e+11 meters, the suns mass is 1.00*10^30kg, what is the orbital speed of the asteroid? Please explain step by step thanks
At first glance, your problem statement is subject to interpretation.
1---Is the 4.44e^11 distance meant to mean
…(4.44e)^11 = 7.9150x10^11 meters
…or 4.44(e^11) = 265,841 meters?
2---Is the distance, r, meant to be from the center of the sun or the surface of the sun?
3—If the center of the sun, the distance can only be 7.9150x10^11 meters.
….If the surface of the sun, either can apply.
4---Providing the mass of the sun implies that you need to compute the gravitational constant of the sun = GM where G = the universal gravitational constant, 6.676259x10^-11m^3/kg.sec^2 and M = the mass of the sun.
5---You give the mass of the sun as 1.00x10^30kg, when, in reality, it is closer to 2.0x10^30 or
1.989157x10^30kg.
5---The actual gravitational constant of the sun is µ = 1.327283x10^20m^3/sec.^2.
Once you have sorted out which of the quantities you intend to use, the velocity required to maintain a circular orbit around the sun may be computed from the following:
Vc = sqrt(µ/r)
thnk you so much.!
To find the orbital speed of the asteroid, we can use the formula for the orbital speed of an object moving in a circular orbit:
v = √(G * M / r)
Where:
- v is the orbital speed
- G is the gravitational constant (approximately 6.67430 × 10^-11 m³⋅kg⁻¹⋅s⁻²)
- M is the mass of the central body (in this case, the mass of the Sun)
- r is the distance between the asteroid and the center of the central body
Given:
- M (mass of the Sun) = 1.00 × 10^30 kg
- r (distance from the asteroid to the Sun) = 4.44 × 10^11 meters
Now, let's plug these values into the formula and calculate the orbital speed:
v = √(G * M / r)
= √((6.67430 × 10^-11 m³⋅kg⁻¹⋅s⁻²) * (1.00 × 10^30 kg) / (4.44 × 10^11 meters))
Simplifying further:
v = √(6.67430 × 10^-11 m³⋅kg⁻¹⋅s⁻² * 1.00 × 10^30 kg / 4.44 × 10^11 meters)
= √(1.00 × 10^19 m³⋅kg⁻¹⋅s⁻² / 4.44 × 10^11 meters)
Now, let's simplify the units:
1 m³⋅kg⁻¹⋅s⁻² / 1 meter = 1 m⋅s⁻²
v = √(1.00 × 10^19 m⋅s⁻² / 4.44 × 10^11 meters)
= √(1.00 × 10^8 s⁻²)
Finally, taking the square root:
v = 1.00 × 10^4 s⁻¹
So the orbital speed of the asteroid is 1.00 × 10^4 meters per second.