<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
Answer:
<u>954.4m/s</u>
Explanation:
For a free falling object,it has constant acceleration and a changing velocity.
By using the velocity-time formula, the velocity can be obtained.
The height the rock travelled is the distance.
From,
Velocity (v) = Distance (d) / Time(t)
v = 3245m/3.4s
v = <u>954.4m/s</u>
That js the answer I got. Hope it's right.
Answer:
magnetic trains works at the principle of repel on of the advantage is that they are fast and dont really need diesel
Answer:
%
Explanation:
The three readings of her gravity calculation is given as
now the mean value of gravity calculation is given as
now the percentage error in her calculation is given as
%
