Use the following numbers for this question.
MEarth 5.98e+24 kg
MJupiter 1.9e+27 kg
MSun 2e+30 kg
Earth-Sun distance 150000000000 m
Jupiter-Sun distance 778000000000 m
rE 6380000 m
rJ 69000000 m
G = 6.67e-11 Nm2kg-2
the acceleration due to gravity at the surface of Jupiter has been found to be 26.6 m/s^2
How fast would a steel sphere be going if you let it drop 20 m from rest, near the surface of Jupiter?
How many Earth years does it take Jupiter to orbit the Sun once, assuming a circular orbit and one Earth year as 365.25 days?
What is the force of gravity that the sun exerts on the Earth?
The sun pulls on the Earth with a huge amount of force. Why does the Earth not fall into the Sun? Explain
never mind on the last parts, just what is the answer to this question using those numbers
How fast would a steel sphere be going if you let it drop 20 m from rest, near the surface of Jupiter?
To find the speed of the steel sphere when dropped from a height of 20 m near the surface of Jupiter, we can use the concept of gravitational potential energy being converted into kinetic energy. The formula for the speed is given by:
v = sqrt(2gh)
Where:
v is the speed of the sphere
g is the acceleration due to gravity at the surface of Jupiter (26.6 m/s^2)
h is the height (20 m)
Plugging in the given values, we get:
v = sqrt(2 * 26.6 * 20) = 18.2 m/s
So, the steel sphere would be going approximately 18.2 m/s when dropped 20 m near the surface of Jupiter.
To find how many Earth years it takes Jupiter to orbit the Sun once, we can use Kepler's third law of planetary motion which relates the orbital period (T) to the semi-major axis (a):
T^2 = (4π^2 / GM) * a^3
Where:
T is the orbital period (unknown)
G is the gravitational constant (6.67e-11 Nm^2/kg^2)
M is the mass of the sun (2e+30 kg)
a is the distance between Jupiter and the sun (778000000000 m)
Rearranging the formula, we can solve for T:
T = sqrt((4π^2 / GM) * a^3)
Plugging in the given values, we have:
T = sqrt((4 * π^2 / (6.67e-11) * (2e+30)) * (778000000000)^3)
Calculating this equation gives us the orbital period of Jupiter around the Sun in seconds. To convert it to Earth years, we divide the result by the number of seconds in an Earth year. Given that one Earth year is 365.25 days and each day has 24 hours and 60 minutes, we have:
T_Earth_years = T / (365.25 * 24 * 60 * 60)
Calculating this equation will give us the number of Earth years it takes for Jupiter to orbit the Sun once.
To find the force of gravity that the Sun exerts on the Earth, we can use Newton's law of universal gravitation. The formula to calculate gravitational force is given by:
F = (G * MSun * MEarth) / d^2
Where:
F is the force of gravity between the Sun and Earth (unknown)
G is the gravitational constant (6.67e-11 Nm^2/kg^2)
MSun is the mass of the Sun (2e+30 kg)
MEarth is the mass of the Earth (5.98e+24 kg)
d is the Earth-Sun distance (150000000000 m)
Plugging in the given values, we have:
F = (6.67e-11 * 2e+30 * 5.98e+24) / (150000000000)^2
Calculating this equation will give us the force of gravity that the Sun exerts on the Earth.
The reason why the Earth does not fall into the Sun is due to its orbital motion. While the Sun's gravity is strong and pulls the Earth towards it, the Earth possesses enough tangential velocity to keep it in a stable orbit around the Sun. This balance between gravitational force and tangential velocity allows the Earth to continuously orbit the Sun without falling into it.