A satellite orbits the earth at an altitude of 700 km on the equatorial plane
of the earth and it revolves in the same direction as the direction of rotation
of the earth. Considering the radius of a spherical earth as 6300 km and the
acceleration due to gravity as 10 m/s', the tangential velocity of the satellite
In the orbit is
ki/s. [round off to 2 decimal places]
m g = m v^2 / r ... v^2 = g r ... v = √(g r) = √[(10 m/s^2) (7E6 m)]
To find the tangential velocity of the satellite, we can use the formula for the circular motion:
v = √(GMe / r)
where
v is the tangential velocity,
G is the universal gravitational constant (approximately 6.674 × 10^-11 N m^2/kg^2),
Me is the mass of the Earth (approximately 5.972 × 10^24 kg),
and r is the radius of the orbit (equal to the sum of the radius of the Earth and the altitude of the satellite).
First, let's calculate the radius of the orbit:
r = radius of the Earth + altitude of the satellite
r = 6300 km + 700 km
r = 7000 km
Converting the radius to meters:
r = 7000 km × 1000 m/km
r = 7,000,000 m
Now we can substitute the values into the formula:
v = √((6.674 × 10^-11 N m^2/kg^2)(5.972 × 10^24 kg) / 7,000,000 m)
Simplifying the equation:
v = √(3.986032 × 10^14 m^3/kg s^2)
Taking the square root:
v ≈ 6.31 × 10^3 m/s
Rounding off to 2 decimal places, the tangential velocity of the satellite in the orbit is 6,310 m/s.
To find the tangential velocity of the satellite orbiting the Earth, we can use the formula:
v = sqrt(G * M / r)
Where:
- v is the tangential velocity
- G is the gravitational constant (approximately 6.67430 x 10^-11 m^3 kg^-1 s^-2)
- M is the mass of the Earth
- r is the distance between the center of the Earth and the satellite
First, we need to find the distance between the satellite and the center of the Earth. Since the altitude of the satellite above the Earth's surface is 700 km and the radius of the Earth is given as 6300 km, the total distance is:
r = 6300 km + 700 km = 7000 km
Converting km to meters:
r = 7,000,000 meters
Next, we can substitute the values into the formula. The mass of the Earth (M) is not given, but we can assume it to be approximately 5.972 x 10^24 kg.
v = sqrt((6.67430 x 10^-11 m^3 kg^-1 s^-2) * (5.972 x 10^24 kg) / (7,000,000 meters))
Calculating:
v = sqrt(4.4313254 x 10^4 m^2 s^-2)
v ≈ 210.55 m/s
Therefore, the tangential velocity of the satellite in orbit is approximately 210.55 m/s.