1 ) If the mass of the Moon is 7.4 x 1022 kg and its radius is 1.74 x 106 m, compute the speed with which an object would have to be fired in order to sail away from it, completely overcoming the Moon’s gravity pull.
2 ) Show that if a satellite orbits very near the surface of a planet with period T, the density (mass
per unit volume) of the planet is p(rho) = 3pi/(GT)^2
(1)
mv²/2 = GmM/R,
v =sqrt(2•G•M/R)
This is escape velocity (it is the speed needed to "break free" from a gravitational field without further propulsion).
The gravitational constant
G =6.67•10^-11 N•m²/kg².
(2)
The volume of the sphere
V=(4•π•R^3)/3
The planet mass
M=ρ•V
Its density
ρ =M/V=3•M/4•π•R³ …..(1)
The period of satellite
T =2•π•r/v
r is the radius of the satellite orbit (given that r≈R)
v is the orbital speed of a body ( satellite) at which it orbits around a more massive body
From m•v²/R =G•m•M/R²
this speed is v =sqrt(g•M•/R)
Then period is
T=2πR /sqrt(G•M/R)
Square this equation
T² =4•π²•R³/G•M,
R³ =G•M•T²/, …. (2)
Subsitute (2) in (1)
ρ = 3•M/4•π•R³ =
=3•M•4•π²/4•π•G•M• T² =
= 3• π /G•T².
It seems to me that it is mistake in your expression 3pi/(GT)^2.
thank you ...so much..
ya i m sorry ...i wrote the wrong expression by mistake
To answer these questions, we'll use some basic principles of physics and mathematics. Let's go step by step:
1) To calculate the speed required to overcome the Moon's gravity pull and sail away from it, we can use the concept of escape velocity. The escape velocity is the minimum speed an object needs to escape from the gravitational pull of a celestial body.
The formula for escape velocity is given by:
v_e = sqrt((2 * G * M) / R)
Where:
v_e is the escape velocity
G is the gravitational constant (approximately 6.67 x 10^-11 N m^2/kg^2)
M is the mass of the Moon
R is the radius of the Moon
Plugging in the values given:
M = 7.4 x 10^22 kg
R = 1.74 x 10^6 m
We can now calculate the escape velocity:
v_e = sqrt((2 * 6.67 x 10^-11 N m^2/kg^2 * 7.4 x 10^22 kg) / (1.74 x 10^6 m))
Calculating the expression inside the square root will give us the answer.
2) To show that the density of the planet (p) is equal to 3pi/(GT)^2 when a satellite orbits very near the planet's surface with a period (T), we can use some formulas from orbital mechanics.
The formula for the period of an object in circular orbit around a planet is given by:
T = 2 * pi * sqrt((R^3) / (G * M))
Where:
T is the period of the satellite
R is the radius of the planet
G is the gravitational constant
M is the mass of the planet
Rearranging this formula will give us the expression for the radius (R):
R = (T^2 * G * M) / (4 * pi^2)
Now, the volume of a sphere is given by:
V = (4/3) * pi * R^3
We can substitute the expression for R into the volume formula:
V = (4/3) * pi * ((T^2 * G * M) / (4 * pi^2))^3
Simplifying this expression will give us the volume (V) in terms of T, G, and M.
The density (p) is defined as the mass (M) divided by the volume (V):
p = M / V
Substituting the expressions for V and M into the density formula, we can calculate the density in terms of T, G, and pi.
p = M / ((4/3) * pi * ((T^2 * G * M) / (4 * pi^2))^3)
By simplifying this equation, we can show that p = 3pi/(GT)^2.
Please note that these calculations involve some mathematical operations, so make sure to double-check the calculations.