A spaceship is on a straight-line path between the Earth and the Moon. At what distance from Earth is the net gravitational pull on the probe from the Earth and the moon zero? Mass of Earth = 6×10^24 kg. Mass of Moon = 7×10^22 kg.
Can someone please confirm if my answer of 4678km is right.
i used the centre of mass equation....
=(6x10^24*0)+(7x10^22*405608)/ 6x10^24+7x10^22
So how bout you? How'd you get to the answer?
i got a similar answer
how did you work it out?
i used both both masses and the distance to the moon
If the distance between the Earth and the moon is around 380,000km, doesn't 4,678km (assuming it is from the Earth) seem a bit too small? Earth is a much larger body than the moon, so the point where the gravity of the moon and the gravity of the Earth equal each-other (ie, cancel out) would be closer to the moon than to the Earth? I tried using GM(earth)M(spaceship)/r^2=GM(moon)M(spaceship)/(380000-r^2)
where G, and M(spaceship) cancel each other out so you can just solve for r, assuming it is the distance from Earth?
To determine the location where the net gravitational pull on the spaceship from the Earth and the Moon is zero, we can use the concept of gravitational force.
The gravitational force between two objects can be calculated using the equation:
F = (G * m1 * m2) / r^2
Where:
- F is the gravitational force
- G is the gravitational constant (approximately 6.67430 × 10^-11 N m^2/kg^2)
- m1 and m2 are the masses of the two objects
- r is the distance between the centers of the two objects
In this case, we have the mass of the Earth (6 × 10^24 kg), the mass of the Moon (7 × 10^22 kg), and we want to find the distance from Earth at which the net gravitational pull is zero.
Let's assume that the distance from Earth to the desired point is x, and the distance from the Moon to the desired point is d (since the spaceship is on a straight-line path between the Earth and the Moon).
To find the point where the net gravitational pull is zero, we need to set up an equation where the gravitational force from the Earth cancels out the gravitational force from the Moon.
The gravitational force from the Earth is given by:
F_Earth = (G * m_Earth * m_probe) / (x)^2
The gravitational force from the Moon is given by:
F_Moon = (G * m_Moon * m_probe) / (d)^2
Since the net gravitational pull is zero, we can set these two forces equal to each other:
F_Earth = F_Moon
(G * m_Earth * m_probe) / (x)^2 = (G * m_Moon * m_probe) / (d)^2
We can cancel out the common factors:
(m_Earth) / (x)^2 = (m_Moon) / (d)^2
Rearranging the equation, we can solve for d:
d^2 = (m_Moon * x^2) / m_Earth
Taking the square root of both sides:
d = sqrt((m_Moon * x^2) / m_Earth)
Now, we have an equation in terms of x and d. To find the value of x, we need the value of d. Let's substitute the known values into the equation.
m_Earth = 6 × 10^24 kg
m_Moon = 7 × 10^22 kg
Let's assume the distance to the Moon is approximately 384,400 km, which we can convert to meters for consistency:
d = 384,400,000 meters
Now, we can solve for x:
x = sqrt((m_Earth * d^2) / m_Moon)
Plugging in the values:
x = sqrt((6 × 10^24 kg * (384,400,000 meters)^2) / (7 × 10^22 kg))
Evaluating this equation, we find:
x ≈ 466,499.63 km
Therefore, the distance from Earth where the net gravitational pull on the spaceship from the Earth and the Moon is zero is approximately 466,499.63 km.
It seems your answer of 4678 km is significantly smaller. Please recheck your calculations.